Trigonometry

For a triangle $ABC$, use the standard convention that side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$. Matching each angle with its opposite side is essential when using the sine and cosine rules.

Area using two sides and the included angle

The usual area formula is

$$\text{Area}=\frac12\times\text{base}\times\text{perpendicular height}.$$

Suppose sides $a$ and $b$ enclose angle $C$. The component of side $b$ perpendicular to side $a$ is $b\sin C$, so

$$\text{Area}=\frac12a(b\sin C)=\frac12ab\sin C.$$

Therefore,

$$\boxed{\text{Area}=\frac12ab\sin C}$$

and, by relabelling,

$$\boxed{\text{Area} =\frac12bc\sin A =\frac12ca\sin B =\frac12ab\sin C.}$$

The angle in the formula must be the angle between the two stated sides.

Worked example

Two sides of a triangle have lengths $7$ and $10$, and the angle between them is $30^\circ$. Its area is

$$\frac12(7)(10)\sin30^\circ =35\cdot\frac12 =\frac{35}{2}.$$

A common mistake is to use an angle that is not between the two given sides.

The sine rule

The three expressions for the area of a triangle are equal:

$$\frac12bc\sin A = \frac12ca\sin B = \frac12ab\sin C.$$

Dividing by $\frac12abc$ gives

$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.$$

Equivalently,

$$\boxed{ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} }.$$

Use the sine rule when a known side and its opposite angle form a complete pair.

Typical situations are:

• two angles and one side, to find another side;
• two sides and an angle opposite one of them, to find another angle.

Place corresponding sides and angles in the same position in each fraction. For example,

$$\frac{a}{\sin A}=\frac{b}{\sin B},$$

not $\frac{a}{\sin B}=\frac{b}{\sin A}$.

Worked example

In a triangle, $A=30^\circ$, $B=45^\circ$, and $a=6$. To find $b$,

$$\frac{b}{\sin45^\circ} = \frac{6}{\sin30^\circ}.$$

Therefore,

$$b = 6\frac{\sin45^\circ}{\sin30^\circ} = 6\frac{\frac{\sqrt2}{2}}{\frac12} = 6\sqrt2.$$

The ambiguous case

The ambiguous case occurs when two sides and a non-included angle are known. This is often called the angle–side–side or SSA case.

Suppose $A$, $a$, and $b$ are known. The sine rule gives

$$\sin B=\frac{b\sin A}{a}.$$

If this value is between $0$ and $1$, there may be two possible angles:

$$B_1=\sin^{-1}\left(\frac{b\sin A}{a}\right)$$

and

$$B_2=180^\circ-B_1.$$

This happens because

$$\sin\theta=\sin(180^\circ-\theta).$$

Each candidate must be checked against

$$A+B<180^\circ.$$

A candidate that makes the angle sum at least $180^\circ$ does not produce a triangle.

For an acute known angle $A$, define the perpendicular height

$$h=b\sin A.$$

The possibilities are:

Condition	Number of triangles
$a<h$	$0$
$a=h$	$1$, a right-angled triangle
$h<a<b$	$2$
$a\geq b$	$1$

If the known angle $A$ is obtuse, its opposite side $a$ must be the longest side. Therefore:

• $a>b$ gives one triangle;
• $a\leq b$ gives no triangle.

The safest exam method is to find both possible sine solutions and test the angle sum rather than relying only on memorised cases.

Worked example

Suppose

$$A=30^\circ,\qquad a=5,\qquad b=5\sqrt3.$$

The sine rule gives

$$\sin B = \frac{b\sin A}{a} = \frac{5\sqrt3\cdot\frac12}{5} = \frac{\sqrt3}{2}.$$

Therefore,

$$B=60^\circ \quad\text{or}\quad B=120^\circ.$$

Both are possible because

$$30^\circ+60^\circ<180^\circ$$

and

$$30^\circ+120^\circ<180^\circ.$$

The two possible third angles are therefore

$$C=90^\circ \quad\text{or}\quad C=30^\circ.$$

The cosine rule

For any triangle,

$$\boxed{a^2=b^2+c^2-2bc\cos A.}$$

The side isolated on the left is opposite the angle used in the cosine term. The other forms are

$$b^2=c^2+a^2-2ca\cos B$$

and

$$c^2=a^2+b^2-2ab\cos C.$$

The cosine rule can be viewed as Pythagoras’ theorem with a correction term. If $A=90^\circ$, then $\cos A=0$, so

$$a^2=b^2+c^2.$$

Use the cosine rule when:

• two sides and their included angle are known, to find the third side;
• all three sides are known, to find an angle.

To find an angle, rearrange:

$$\boxed{ \cos A=\frac{b^2+c^2-a^2}{2bc}. }$$

Worked example

Two sides have lengths $7$ and $8$, and the included angle is $60^\circ$. If the opposite side is $a$, then

$$a^2 = 7^2+8^2-2(7)(8)\cos60^\circ.$$

Hence

$$a^2 = 49+64-112\left(\frac12\right) = 57,$$

so

$$a=\sqrt{57}.$$

The cosine rule also provides a useful check:

$$\begin{aligned} a^2&<b^2+c^2 &&\Longleftrightarrow A\text{ is acute},\\ a^2&=b^2+c^2 &&\Longleftrightarrow A=90^\circ,\\ a^2&>b^2+c^2 &&\Longleftrightarrow A\text{ is obtuse}. \end{aligned}$$

The longest side must be opposite the largest angle.

Choosing an efficient method

Before calculating, identify the information pattern:

Information given	Efficient method
Two sides and their included angle	Cosine rule for the third side
Three sides	Cosine rule for an angle
A side–opposite-angle pair and another side or angle	Sine rule
Two sides and their included angle, with area required	$\frac12ab\sin C$
Right-angled triangle	Pythagoras and basic right-angle trigonometry are usually faster

The triangle inequality can sometimes avoid unnecessary trigonometry. Three positive lengths form a non-degenerate triangle only when the sum of every pair exceeds the third side. It is enough to check that the two shorter sides add to more than the longest side.

Problems in three dimensions

A three-dimensional problem should usually be reduced to one or more ordinary two-dimensional triangles.

A reliable method is:

1. Mark the required angle or length clearly.
2. Find useful face diagonals using Pythagoras’ theorem.
3. Identify a plane containing the required points.
4. Work only within the resulting triangle.
5. Apply the sine rule, cosine rule, area formula or right-angle trigonometry as appropriate.

Do not use an angle from a flat projection unless it is genuinely the required spatial angle.

For example, in the cuboid shown,

$$AG=\sqrt{3^2+4^2+12^2}=13$$

and

$$BG=\sqrt{4^2+12^2}=4\sqrt{10}.$$

In triangle $ABG$, the cosine rule gives

$$\cos\angle BAG = \frac{AB^2+AG^2-BG^2}{2(AB)(AG)}.$$

Therefore,

$$\cos\angle BAG = \frac{3^2+13^2-(4\sqrt{10})^2}{2(3)(13)} = \frac{18}{78} = \frac{3}{13}.$$

Common traps include pairing the wrong side with an angle, using a non-included angle in the area formula, overlooking the second SSA triangle, and rounding intermediate values unnecessarily.

Degrees divide one complete revolution into $360$ equal parts. Radians instead measure an angle by comparing an arc length with the radius of its circle.

An angle of $1$ radian subtends an arc whose length equals the radius.

More generally,

$$\boxed{\theta=\frac{\text{arc length}}{\text{radius}}}$$

when $\theta$ is measured in radians.

Since the circumference of a circle is $2\pi r$, one complete revolution is

$$2\pi\text{ radians}.$$

Therefore,

$$360^\circ=2\pi, \qquad 180^\circ=\pi, \qquad 90^\circ=\frac{\pi}{2}.$$

Converting between degrees and radians

To convert degrees to radians, multiply by $\frac{\pi}{180}$:

$$\boxed{ \theta^\circ=\frac{\theta\pi}{180}\text{ radians}. }$$

To convert radians to degrees, multiply by $\frac{180}{\pi}$:

$$\boxed{ \theta\text{ radians}=\frac{180\theta}{\pi}^\circ. }$$

Important conversions are:

Degrees	Radians
$0^\circ$	$0$
$30^\circ$	$\frac{\pi}{6}$
$45^\circ$	$\frac{\pi}{4}$
$60^\circ$	$\frac{\pi}{3}$
$90^\circ$	$\frac{\pi}{2}$
$180^\circ$	$\pi$
$270^\circ$	$\frac{3\pi}{2}$
$360^\circ$	$2\pi$

An angle written without a degree symbol is normally interpreted as being in radians.

Arc length

For a central angle $\theta$ radians in a circle of radius $r$,

$$\boxed{s=r\theta.}$$

This follows immediately from the definition

$$\theta=\frac{s}{r}.$$

The formula $s=r\theta$ requires $\theta$ to be in radians.

Area of a sector

A sector with angle $\theta$ is the fraction $\frac{\theta}{2\pi}$ of a full circle. Therefore,

$$\text{Area of sector} = \frac{\theta}{2\pi}\cdot\pi r^2,$$

so

$$\boxed{ \text{Area of sector}=\frac12r^2\theta. }$$

Area of a segment

A segment is the region between a chord and its corresponding arc. For a minor segment with $0<\theta<\pi$,

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}.$$

The two radii form a triangle with sides $r$ and $r$ and included angle $\theta$, so

$$\text{Area of triangle} = \frac12r^2\sin\theta.$$

Therefore,

$$\boxed{ \text{Area of minor segment} = \frac12r^2(\theta-\sin\theta). }$$

The angle $\theta$ must be in radians in the sector term.

The major segment can be found by subtracting the minor segment from the full circle:

$$\text{Area of major segment} = \pi r^2-\text{Area of minor segment}.$$

The chord bounding the segment has length

$$\boxed{ 2r\sin\left(\frac{\theta}{2}\right). }$$

This follows by bisecting the isosceles triangle formed by the two radii.

Worked example

A circle has radius $6$, and a sector has angle $\frac{2\pi}{3}$.

Its arc length is

$$s = 6\left(\frac{2\pi}{3}\right) = 4\pi.$$

Its sector area is

$$\frac12(6^2)\left(\frac{2\pi}{3}\right) = 12\pi.$$

The triangle inside the sector has area

$$\frac12(6^2)\sin\left(\frac{2\pi}{3}\right) = 18\cdot\frac{\sqrt3}{2} = 9\sqrt3.$$

Therefore, the minor segment area is

$$12\pi-9\sqrt3.$$

Common mistakes include using degrees directly in $s=r\theta$, forgetting the factor $\frac12$ in the sector formula, and giving the sector area when the question asks for a segment.

The exact trigonometric values for $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, and $90^\circ$ must be recalled quickly and without a calculator.

They can be reconstructed from two standard triangles.

The $45^\circ$-$45^\circ$-$90^\circ$ triangle

Take an isosceles right-angled triangle whose shorter sides both have length $1$. Pythagoras’ theorem gives the hypotenuse:

$$\sqrt{1^2+1^2}=\sqrt2.$$

Therefore,

$$\sin45^\circ = \cos45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2},$$

and

$$\tan45^\circ=1.$$

The $30^\circ$-$60^\circ$-$90^\circ$ triangle

Bisect an equilateral triangle of side length $2$. Each resulting right-angled triangle has:

• hypotenuse $2$;
• shorter side $1$;
• remaining side $\sqrt3$.

This gives

$$\sin30^\circ=\frac12, \qquad \cos30^\circ=\frac{\sqrt3}{2}, \qquad \tan30^\circ=\frac{1}{\sqrt3}.$$

It also gives

$$\sin60^\circ=\frac{\sqrt3}{2}, \qquad \cos60^\circ=\frac12, \qquad \tan60^\circ=\sqrt3.$$

Exact-value table

Angle	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
Radians	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\sin\theta$	$0$	$\frac12$	$\frac{\sqrt2}{2}$	$\frac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\frac{\sqrt3}{2}$	$\frac{\sqrt2}{2}$	$\frac12$	$0$
$\tan\theta$	$0$	$\frac{1}{\sqrt3}$	$1$	$\sqrt3$	undefined

A quick memory pattern for sine is

$$\sin\theta = \frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}, \frac{\sqrt4}{2}.$$

The cosine values are the same list in reverse order. Tangent can then be obtained from

$$\tan\theta=\frac{\sin\theta}{\cos\theta}.$$

Tangent is undefined at $90^\circ$ because this would require division by

$$\cos90^\circ=0.$$

Values outside the first quadrant can be found from the reference angle and the sign of the function. For example,

$$\sin150^\circ = \sin30^\circ = \frac12,$$

$$\cos210^\circ = -\cos30^\circ = -\frac{\sqrt3}{2},$$

and

$$\tan315^\circ = -\tan45^\circ = -1.$$

Keep exact values in surd form. Replacing $\frac{\sqrt3}{2}$ with a decimal usually makes comparisons and later algebra less efficient.

Sine and cosine can be understood using a point moving around a unit circle. If the point is at angle $x$ from the positive horizontal axis, its coordinates are

$$(\cos x,\sin x).$$

Thus, cosine is the horizontal projection and sine is the vertical projection. Tangent is

$$\tan x=\frac{\sin x}{\cos x},$$

so it represents the ratio of the vertical projection to the horizontal projection whenever $\cos x\neq0$.

The sine graph

The graph $y=\sin x$:

• passes through $(0,0)$;
• has maximum value $1$;
• has minimum value $-1$;
• has period $2\pi$, or $360^\circ$;
• has range $[-1,1]$.

Its key values over one period are

$$\begin{array}{c|ccccc} x&0&\frac{\pi}{2}&\pi&\frac{3\pi}{2}&2\pi\\ \hline \sin x&0&1&0&-1&0 \end{array}$$

Its zeros occur at

$$x=k\pi,\qquad k\in\mathbb Z.$$

The cosine graph

The graph $y=\cos x$:

• begins at $(0,1)$;
• has maximum value $1$;
• has minimum value $-1$;
• has period $2\pi$, or $360^\circ$;
• has range $[-1,1]$.

Its key values over one period are

$$\begin{array}{c|ccccc} x&0&\frac{\pi}{2}&\pi&\frac{3\pi}{2}&2\pi\\ \hline \cos x&1&0&-1&0&1 \end{array}$$

Its zeros occur at

$$x=\frac{\pi}{2}+k\pi,\qquad k\in\mathbb Z.$$

The tangent graph

The graph $y=\tan x$:

• passes through $(0,0)$;
• has period $\pi$, or $180^\circ$;
• has range $\mathbb R$;
• is undefined where $\cos x=0$;
• has vertical asymptotes at

$$x=\frac{\pi}{2}+k\pi,\qquad k\in\mathbb Z.$$

Its zeros occur at

$$x=k\pi,\qquad k\in\mathbb Z.$$

Periodicity

A periodic function repeats after a fixed horizontal interval.

For sine and cosine,

$$\sin(x+2\pi)=\sin x$$

and

$$\cos(x+2\pi)=\cos x.$$

For tangent,

$$\tan(x+\pi)=\tan x.$$

More generally,

$$\sin(x+2k\pi)=\sin x,$$

$$\cos(x+2k\pi)=\cos x,$$

and

$$\tan(x+k\pi)=\tan x,$$

where $k$ is any integer.

Symmetry

Sine is an odd function:

$$\boxed{\sin(-x)=-\sin x.}$$

Its graph has rotational symmetry of $180^\circ$ about the origin.

Cosine is an even function:

$$\boxed{\cos(-x)=\cos x.}$$

Its graph has reflection symmetry in the $y$-axis.

Tangent is an odd function:

$$\boxed{\tan(-x)=-\tan x.}$$

Its graph has rotational symmetry of $180^\circ$ about the origin.

Other useful symmetry relationships include

$$\sin(\pi-x)=\sin x,$$

$$\cos(\pi-x)=-\cos x,$$

and

$$\tan(\pi-x)=-\tan x.$$

In degrees, replace $\pi$ by $180^\circ$.

Signs in the four quadrants

The signs follow from the coordinates $(\cos x,\sin x)$:

Quadrant	Positive functions
I	sine, cosine and tangent
II	sine only
III	tangent only
IV	cosine only

Transformations and periods

For

$$y=A\sin(Bx+C)+D$$

or

$$y=A\cos(Bx+C)+D,$$

the amplitude is

$$|A|,$$

the period in radians is

$$\boxed{\frac{2\pi}{|B|}},$$

and the central horizontal line is

$$y=D.$$

The horizontal translation is

$$-\frac{C}{B}.$$

It is often safest to factor the expression inside the function:

$$2x+\frac{\pi}{3} = 2\left(x+\frac{\pi}{6}\right).$$

Therefore, the horizontal shift is $\frac{\pi}{6}$ to the left, not $\frac{\pi}{3}$.

For

$$y=A\tan(Bx+C)+D,$$

the period is

$$\boxed{\frac{\pi}{|B|}}.$$

Tangent has no amplitude because it is unbounded.

In degrees, the corresponding periods are

$$\frac{360^\circ}{|B|}$$

for sine and cosine, and

$$\frac{180^\circ}{|B|}$$

for tangent.

Graphs can often answer comparison or solution-count questions faster than lengthy algebra. For example, on $0<x<\frac{\pi}{2}$, sine is increasing and cosine is decreasing. This allows exact values to be ordered without calculating unfamiliar expressions.

The two identities required are

$$\boxed{\tan\theta=\frac{\sin\theta}{\cos\theta}}$$

and

$$\boxed{\sin^2\theta+\cos^2\theta=1.}$$

Here,

$$\sin^2\theta$$

means

$$(\sin\theta)^2.$$

It does not mean $\sin(\theta^2)$.

Deriving the Pythagorean identity

On the unit circle, the point at angle $\theta$ has coordinates

$$(\cos\theta,\sin\theta).$$

Every point on the unit circle satisfies

$$x^2+y^2=1.$$

Substituting $x=\cos\theta$ and $y=\sin\theta$ gives

$$\cos^2\theta+\sin^2\theta=1.$$

This is Pythagoras’ theorem expressed using trigonometric coordinates, and it is valid for every real angle.

Useful rearrangements are

$$\boxed{\sin^2\theta=1-\cos^2\theta}$$

and

$$\boxed{\cos^2\theta=1-\sin^2\theta.}$$

These allow equations involving both sine and cosine to be converted into equations involving only one function.

Deriving the tangent identity

In a right-angled triangle,

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$

and

$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}.$$

Therefore,

$$\frac{\sin\theta}{\cos\theta} = \frac{\frac{\text{opposite}}{\text{hypotenuse}}} {\frac{\text{adjacent}}{\text{hypotenuse}}} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta.$$

The identity is valid only when

$$\cos\theta\neq0,$$

because tangent is undefined when the denominator is zero.

Using the identities

For example,

$$\frac{1-\sin^2\theta}{\cos\theta}$$

can be simplified using

$$1-\sin^2\theta=\cos^2\theta.$$

Thus,

$$\frac{1-\sin^2\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta,$$

provided that $\cos\theta\neq0$.

The restriction matters: cancelling a factor of $\cos\theta$ silently assumes it is non-zero.

The Pythagorean identity can also be used to compare expressions. For example,

$$3\sin^2\theta+3\cos^2\theta = 3(\sin^2\theta+\cos^2\theta) = 3.$$

Useful algebraic habits include:

• converting everything to sine when an equation is quadratic in $\sin\theta$;
• converting everything to cosine when it is quadratic in $\cos\theta$;
• replacing tangent by $\frac{\sin\theta}{\cos\theta}$ when this exposes a factorisation;
• avoiding division by a trigonometric expression unless its zero cases have already been considered.

Also remember that

$$\sqrt{\sin^2\theta}=|\sin\theta|,$$

not necessarily $\sin\theta$. Taking the square root of a squared equation usually introduces both positive and negative cases.

A trigonometric equation normally has infinitely many solutions because the functions are periodic. An interval restricts the answer to a finite set.

A reliable method is:

1. Check whether angles are measured in degrees or radians.
2. Rearrange until one trigonometric expression is isolated, or convert the equation into one trigonometric function.
3. Find the reference or principal solution.
4. Use symmetry and periodicity to generate every solution for the whole inner angle.
5. Rearrange to find $x$.
6. Keep only values in the stated interval.
7. Check excluded endpoints and any values lost through division or square roots.

When the equation contains $ax+b$, it is often efficient to set

$$y=ax+b.$$

Convert the interval for $x$ into an interval for $y$, solve for $y$, and only then return to $x$. This prevents solutions from being missed.

Standard solution forms

If

$$\sin y=s,$$

let

$$\alpha=\sin^{-1}s.$$

Then

$$\boxed{ y=\alpha+2k\pi \quad\text{or}\quad y=\pi-\alpha+2k\pi, }$$

where $k\in\mathbb Z$.

If

$$\cos y=c,$$

let

$$\beta=\cos^{-1}c.$$

Then

$$\boxed{ y=\pm\beta+2k\pi. }$$

If

$$\tan y=t,$$

let

$$\gamma=\tan^{-1}t.$$

Then

$$\boxed{ y=\gamma+k\pi. }$$

In degrees, replace $2\pi$ by $360^\circ$ and $\pi$ by $180^\circ$.

These formulas are useful, but a correctly labelled graph or quadrant diagram is equally valid and can be faster for exact-value equations.

Worked example: tangent equation

Solve

$$\tan x=-\frac{1}{\sqrt3}$$

for

$$-\pi<x<\pi.$$

The reference angle is $\frac{\pi}{6}$, because

$$\tan\frac{\pi}{6}=\frac{1}{\sqrt3}.$$

Tangent is negative in quadrants II and IV. Its general solution is

$$x=-\frac{\pi}{6}+k\pi.$$

The values in the interval are

$$\boxed{ x=-\frac{\pi}{6},\quad \frac{5\pi}{6}. }$$

Equations involving a square

If

$$\sin^2 y=\frac12,$$

then

$$\sin y=\pm\frac{1}{\sqrt2}.$$

Both signs are required. Equivalently, the solutions can be written compactly as

$$y=\frac{\pi}{4}+\frac{k\pi}{2}.$$

Worked example: transformed squared equation

Solve

$$\sin^2\left(2x+\frac{\pi}{3}\right)=\frac12$$

for

$$-2\pi<x<2\pi.$$

Let

$$y=2x+\frac{\pi}{3}.$$

Since $\sin^2y=\frac12$,

$$y=\frac{\pi}{4}+\frac{k\pi}{2}.$$

Therefore,

$$2x+\frac{\pi}{3} = \frac{\pi}{4}+\frac{k\pi}{2},$$

so

$$x = -\frac{\pi}{24}+\frac{k\pi}{4}.$$

Applying the interval gives

$$-2\pi < -\frac{\pi}{24}+\frac{k\pi}{4} < 2\pi.$$

This is satisfied by

$$k=-7,-6,\ldots,8.$$

Hence the complete solution set is

$$\boxed{ x=-\frac{\pi}{24}+\frac{k\pi}{4}, \qquad k\in\{-7,-6,\ldots,8\}. }$$

There are $16$ solutions.

Equations involving both sine and cosine

Use

$$\sin^2x+\cos^2x=1$$

to rewrite the equation using only one trigonometric function.

Worked example

Solve

$$12\cos^2x+6\sin x-10=2$$

for

$$0^\circ<x<360^\circ.$$

Use

$$\cos^2x=1-\sin^2x:$$

$$12(1-\sin^2x)+6\sin x-10=2.$$

Simplifying gives

$$-12\sin^2x+6\sin x=0.$$

Factorising,

$$6\sin x(1-2\sin x)=0.$$

Therefore,

$$\sin x=0$$

or

$$\sin x=\frac12.$$

Within the stated open interval,

$$\sin x=0$$

gives

$$x=180^\circ,$$

while

$$\sin x=\frac12$$

gives

$$x=30^\circ,\ 150^\circ.$$

Thus,

$$\boxed{x=30^\circ,\ 150^\circ,\ 180^\circ.}$$

Do not divide the equation by $\sin x$, because that would lose the valid solution arising from $\sin x=0$.

Counting solutions graphically

Some questions require only the number of solutions. Sketching both sides as functions can be faster than finding every value explicitly.

For an equation

$$f(x)=g(x),$$

the number of solutions is the number of intersections of the graphs

$$y=f(x) \quad\text{and}\quad y=g(x)$$

within the interval.

Use the period to estimate how many repeating sections occur, but inspect the endpoints and turning points carefully. A horizontal line may:

• miss a sine or cosine graph;
• touch it once at a maximum or minimum;
• cross it twice in one period.

Since

$$-1\leq\sin x\leq1$$

and

$$-1\leq\cos x\leq1,$$

equations such as

$$\sin x=2$$

or

$$\cos x=-\frac32$$

have no real solutions.

Tangent can take every real value, although its asymptotes divide the graph into separate branches.

The most frequent errors are:

• mixing degrees and radians;
• finding only the principal solution;
• forgetting the second sine or cosine solution;
• forgetting the negative square root after removing a square;
• solving for $x$ before generating all solutions for the inner angle;
• including an endpoint that the interval excludes;
• dividing by $\sin x$, $\cos x$, or another expression that could be zero;
• accepting an inverse-trigonometric output without checking the correct quadrant;
• rounding exact values too early.

A final substitution or graphical check is often quick and can detect an omitted or extraneous solution.