6.10 - Solving trigonometric equations

6.10 - Solving trigonometric equations

Solve equations involving a single angle by combining algebra, reference angles and the graphs you already know. Every solution must satisfy both the original equation and its stated interval. All angles here are in degrees.

All the solutions for sine and cosine

A calculator gives 30∘30^\circ for sin⁡−1(12)\sin^{-1}(\frac12). Why can sin⁡2x=14\sin^2x=\frac14 have four solutions in one full turn? We will build the answer by first solving equations without a square.

You already know

On the unit circle, sine is the vertical coordinate and cosine is the horizontal coordinate. Reference angles supply magnitudes; quadrants supply signs. Sine and cosine lie between −1-1 and 11, and each repeats after 360∘360^\circ.

An inverse-trigonometric button returns one angle, not a complete solution set. For example, sin⁡x=12\sin x=\frac12 means a unit-circle height of 12\frac12. Both the first and second quadrants reach that height. The graph meets the horizontal line twice.

[DIAGRAM: asset_name: Sine and cosine equation intersections; asset_slug: l2fm-sine-cosine-equation-roots; caption: A horizontal line at one half cuts sine and cosine twice in a full turn; description: Two degree-labelled graphs run from 0 to 360 degrees. The sine graph meets y equals one half at 30 and 150 degrees. The cosine graph meets y equals one half at 60 and 300 degrees. Filled points and vertical dashed guides identify all four intersections in their respective graphs.]
A horizontal line at one half cuts sine and cosine twice in a full turn

For a reference angle α\alpha strictly between 0∘0^\circ and 90∘90^\circ, the following arrangements collect the quadrant information in one place. Axis values need a separate check because two arrangements can coincide there.

Reference-angle arrangements in one full turn

For 0∘<α<90∘0^\circ<\alpha<90^\circ and 0∘⩽x⩽360∘0^\circ\leqslant x\leqslant360^\circ,

equationsolutionssin⁡x=sin⁡αα, 180∘−αsin⁡x=−sin⁡α180∘+α, 360∘−αcos⁡x=cos⁡αα, 360∘−αcos⁡x=−cos⁡α180∘−α, 180∘+α\begin{array}{c|c} \text{equation}&\text{solutions}\\ \hline \sin x=\sin\alpha&\alpha,\ 180^\circ-\alpha\\ \sin x=-\sin\alpha&180^\circ+\alpha,\ 360^\circ-\alpha\\ \cos x=\cos\alpha&\alpha,\ 360^\circ-\alpha\\ \cos x=-\cos\alpha&180^\circ-\alpha,\ 180^\circ+\alpha \end{array}

These arrangements follow from unit-circle symmetry. Exact trigonometric values are also needed when an exact answer is required.
Not on the formulae sheet: learn it.

First isolate the ratio. Then use its sign to select quadrants, work out candidate angles and apply the stated interval. With a calculator, select degree mode and retain extra digits until the final answer.

Worked example 1: two equal sine heights

Question. Solve sin⁡x=12\sin x=\frac12 for 0∘⩽x⩽360∘0^\circ\leqslant x\leqslant360^\circ.

Plan. Use the exact reference angle and choose the two quadrants where sine is positive.

α=30∘because sin⁡30∘=12x=αor180∘−αfirst and second quadrantsx=30∘or150∘both are in the interval\begin{aligned} \alpha&=30^\circ && \text{because }\sin30^\circ=\tfrac12\\ x&=\alpha\quad\text{or}\quad180^\circ-\alpha && \text{first and second quadrants}\\ x&=30^\circ\quad\text{or}\quad150^\circ && \text{both are in the interval} \end{aligned}

Why this step? The same positive height occurs on both sides of the unit circle, so the inverse-sine answer alone is incomplete.

Answer. x=30∘,150∘x=30^\circ,150^\circ.

Check. Both angles have sine 12\frac12. Neither endpoint has that sine value.

Worked example 2 changes only the function from sine to cosine. The target value and interval stay the same, but now the horizontal coordinate must be positive.

Worked example 2: two equal cosine coordinates

Question. Solve cos⁡x=12\cos x=\frac12 for 0∘⩽x⩽360∘0^\circ\leqslant x\leqslant360^\circ.

Plan. Use the cosine reference angle and the first and fourth quadrants.

α=60∘because cos⁡60∘=12x=αor360∘−αcosine positivex=60∘or300∘both are in the interval\begin{aligned} \alpha&=60^\circ && \text{because }\cos60^\circ=\tfrac12\\ x&=\alpha\quad\text{or}\quad360^\circ-\alpha && \text{cosine positive}\\ x&=60^\circ\quad\text{or}\quad300^\circ && \text{both are in the interval} \end{aligned}

Why this step? Sine and cosine have different reference angles for the same value, and their positive quadrants differ.

Answer. x=60∘,300∘x=60^\circ,300^\circ.

Check. Reflection in the horizontal axis keeps the horizontal coordinate 12\frac12 unchanged.

For negative targets, change the allowed quadrants: sin⁡x=−12\sin x=-\frac12 gives 210∘,330∘210^\circ,330^\circ, whereas cos⁡x=−12\cos x=-\frac12 gives 120∘,240∘120^\circ,240^\circ, in the same closed full-turn interval.

Boundary values deserve their own check. In 0∘⩽x⩽360∘0^\circ\leqslant x\leqslant360^\circ, sin⁡x=1\sin x=1 has only 90∘90^\circ, while sin⁡x=0\sin x=0 has 0∘,180∘,360∘0^\circ,180^\circ,360^\circ. The equation cos⁡x=1\cos x=1 has both 0∘0^\circ and 360∘360^\circ: they are different allowed angle values. A target outside [−1,1][-1,1], such as cos⁡x=32\cos x=\frac32, gives no solution.

Watch out

Wrong: always add 180∘180^\circ to get the second sine solution. For sin⁡x=12\sin x=\frac12, 210∘210^\circ has the wrong sign. Right: use the graph or quadrant signs, giving 180∘−30∘=150∘180^\circ-30^\circ=150^\circ.

Isolate cosine in this new equation, then use the negative sign and an exact reference angle.

Tangent solutions and the interval filter

Tangent repeats after a half-turn. A line through the origin meets opposite points of the unit circle with the same coordinate quotient: both numerator and denominator change sign together. That is why its solution spacing differs from sine's usual reflection pattern.

The half-turn repetition of tangent

tan⁡(x+180∘)=tan⁡x\tan(x+180^\circ)=\tan x

This holds wherever tangent is defined. It is undefined when cosine is zero, including 90∘90^\circ and 270∘270^\circ in a full turn.
Not on the formulae sheet: learn it.

For a negative tangent target, use a positive acute reference angle, then choose the second and fourth quadrants. If the question gives a shorter interval, discard candidates outside it after generating them.

Worked example 3: a non-exact tangent value

Question. Solve tan⁡x=−2\tan x=-2 for 0∘⩽x⩽360∘0^\circ\leqslant x\leqslant360^\circ. Give answers to 11 decimal place.

Plan. Work out the acute reference angle using tan⁡−12\tan^{-1}2, then place it in the negative-tangent quadrants.

α=tan⁡−12=63.4349…∘positive reference anglex=180∘−α=116.5650…∘second quadrantorx=360∘−α=296.5650…∘fourth quadrant\begin{aligned} \alpha&=\tan^{-1}2=63.4349\ldots^\circ && \text{positive reference angle}\\ x&=180^\circ-\alpha=116.5650\ldots^\circ && \text{second quadrant}\\ \text{or}\quad x&=360^\circ-\alpha=296.5650\ldots^\circ && \text{fourth quadrant} \end{aligned}

Why this step? A negative calculator output such as −63.4349…∘-63.4349\ldots^\circ does not lie in the stated interval; the quadrant construction gives the required representatives.

Answer. x=116.6∘,296.6∘x=116.6^\circ,296.6^\circ.

Check. The unrounded angles differ by exactly 180∘180^\circ, and both give tangent −2-2. Rounded answers give values close to −2-2.

Worked example 4 changes only the interval. The equation and its candidate angles stay unchanged; one candidate now fails the interval test.

Worked example 4: a shorter interval

Question. Solve tan⁡x=−2\tan x=-2 for 100∘<x⩽280∘100^\circ<x\leqslant280^\circ. Give answers to 11 decimal place.

Plan. Generate the same full-turn candidates, then compare them with the new endpoints before rounding.

α=tan⁡−12=63.4349…∘same reference anglex=116.5650…∘or296.5650…∘same two candidates100∘<116.5650…∘⩽280∘keep the first296.5650…∘>280∘discard the second\begin{aligned} \alpha&=\tan^{-1}2=63.4349\ldots^\circ && \text{same reference angle}\\ x&=116.5650\ldots^\circ\quad\text{or}\quad296.5650\ldots^\circ && \text{same two candidates}\\ 100^\circ&<116.5650\ldots^\circ\leqslant280^\circ && \text{keep the first}\\ 296.5650\ldots^\circ&>280^\circ && \text{discard the second} \end{aligned}

Why this step? Satisfying the equation is necessary, but a solution must also be an angle the question allows.

Answer. x=116.6∘x=116.6^\circ only.

Check. The retained unrounded angle gives tangent −2-2 and lies strictly above 100∘100^\circ, below 280∘280^\circ.

The same idea works for intervals that cross a full-turn boundary: use the graph's period to move candidates into the required window. Read strict and inclusive endpoints literally. For instance, 360∘360^\circ is allowed by x⩽360∘x\leqslant360^\circ, but excluded by x<360∘x<360^\circ.

Watch out

Wrong: accept 90∘90^\circ because it lies in the requested interval. Right: check the original function too. Tangent is undefined there, so it cannot solve a tangent equation.

Change the target and interval, keeping all calculator digits until the final line.

A square creates two possible signs

If u2=14u^2=\frac14, then u=12u=\frac12 or u=−12u=-\frac12. Replacing uu by a trigonometric ratio does not remove either sign. Solve both resulting equations before collecting the angles.

Worked example 5: four solutions from one square

Question. Solve sin⁡2x=14\sin^2x=\frac14 for 0∘⩽x⩽360∘0^\circ\leqslant x\leqslant360^\circ.

Plan. Take both square roots, then solve the positive-sine and negative-sine branches.

sin⁡x=±12both square rootssin⁡x=12  ⟹  x=30∘,150∘positive sinesin⁡x=−12  ⟹  x=210∘,330∘negative sine\begin{aligned} \sin x&=\pm\frac12 && \text{both square roots}\\ \sin x=\frac12&\implies x=30^\circ,150^\circ && \text{positive sine}\\ \sin x=-\frac12&\implies x=210^\circ,330^\circ && \text{negative sine} \end{aligned}

Why this step? Squaring hides the sign, so an equation with a squared ratio can have more solutions than either unsquared branch.

Answer. x=30∘,150∘,210∘,330∘x=30^\circ,150^\circ,210^\circ,330^\circ.

Check. The four sine values are 12,12,−12,−12\frac12,\frac12,-\frac12,-\frac12. Squaring any of them gives 14\frac14.

Here is the opening puzzle's answer: the calculator returned one angle from just one sign branch. Both signs together give four. There is another neat consequence: sine takes a full turn to repeat, but sine squared repeats after a half-turn, because changing the sign leaves the square unchanged.

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