Solving Basic Trigonometric Equations

Video lesson · from 11:204 worked examples

One lesson video covers all of Trigonometry, so it opens at 11:20 for this topic — not from the beginning.

To solve basic trigonometric equations:

Rearrange the equation into the form $\sin x = k$ , $\cos x = k$ or $\tan x = k$ .
Find the base angle (the related acute angle in the first quadrant) by taking the inverse trig function of the positive value of $k$ .
Use the CAST diagram to determine which two quadrants the solutions lie in based on the sign of $k$ .
Calculate your final answers using the base angle and the quadrants, checking that they fall inside the required domain.

Give every solution in the domain: a calculator returns one angle. Each equation normally has two solutions per revolution — use CAST to find the others.
Check the domain and its units: $0 \leq x \leq 360$ wants degrees; $0 \leq x \leq 2\pi$ wants radians. Answering in the wrong one loses the marks.
Isolate the trig function first: rearrange to $\sin x = k$ before going near a calculator.
A negative value does not mean a negative angle: it tells you which quadrants to use. Your answers should still sit inside the given domain.

Example 1

Solve $\sin x^\circ = \frac{1}{2}$ for $0 < x^\circ < 360$ .

Base angle: $\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$

Since Sine is positive, solutions are in quadrants 1 and 2.

\begin{aligned} x &= 30 \quad \text{(Q1)} \\ x &= 180 - 30 \\ &= 150 \quad \text{(Q2)} \end{aligned}

Solutions: 30°, 150°

Example 2

Solve $\cos x = -\frac{1}{\sqrt{5}}$ for $0 < x < 2\pi$ .

Base angle: $\cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \approx 1.107 \text{ rad}$

Since Cosine is negative, solutions are in quadrants 2 and 3.

\begin{aligned} x &= \pi - 1.107 \\ &\approx 2.034 \quad \text{(Q2)} \\ x &= \pi + 1.107 \\ &\approx 4.249 \quad \text{(Q3)} \end{aligned}

Solutions: 2.034, 4.249

Example 3

Solve $\sin x = 3$ for $0 < x < 2\pi$ .

The maximum value of the sine function is 1 and the minimum is -1.

Therefore, there are no solutions because $-1 \leq \sin x \leq 1$ .

Example 4

Solve $\tan x^\circ = -5$ for $0 < x^\circ < 720$ .

Base angle: $\tan^{-1}(5) \approx 78.7^\circ$

Since Tangent is negative, solutions are in quadrants 2 and 4.

Because the domain is up to $720^\circ$ , we need to find two cycles of solutions by adding $360^\circ$ to the original solutions.

\begin{aligned} x &= 180 - 78.7 \\ &= 101.3 \quad \text{(First cycle, Q2)} \\ x &= 360 - 78.7 \\ &= 281.3 \quad \text{(First cycle, Q4)} \\ x &= 101.3 + 360 \\ &= 461.3 \quad \text{(Second cycle, Q2)} \\ x &= 281.3 + 360 \\ &= 641.3 \quad \text{(Second cycle, Q4)} \end{aligned}