Trigonometric Equations with Square Terms

Video lesson · from 22:312 worked examples

One lesson video covers all of Trigonometry, so it opens at 22:31 for this topic — not from the beginning.

Theory

When solving an equation involving a squared term (like $\sin^2 x$ or $\cos^2 x$ ), taking the square root requires remembering the plus or minus ( $\pm$ ) sign.

This means you will usually be looking for solutions in all four quadrants.

⚠️ Common Examiner Traps

Do not lose the negative root: $\sin^2 x = \tfrac{1}{4}$ gives $\sin x = \pm\tfrac{1}{2}$ . Taking only the positive root halves your solutions.
That means four solutions, not two: each of the two values produces its own pair of angles.
Isolate the squared term first: rearrange fully before square rooting.
Use CAST for each value separately: the positive and negative cases sit in different quadrants.

Worked examples

Example 1

Solve $\tan^2 x^\circ = 3$ for $0 < x^\circ < 360$ .

Take the square root of both sides:

\tan x^\circ = \pm\sqrt{3}

Base angle: $\tan^{-1}(\sqrt{3}) = 60^\circ$

Since we have both positive and negative values, there will be solutions in all four quadrants.

\begin{aligned} x &= 60 \quad \text{(Q1)} \\ x &= 180 - 60 \\ &= 120 \quad \text{(Q2)} \\ x &= 180 + 60 \\ &= 240 \quad \text{(Q3)} \\ x &= 360 - 60 \\ &= 300 \quad \text{(Q4)} \end{aligned}

Example 2

Solve $4\cos^2 x = 3$ for $0 < x < 2\pi$ .

\begin{aligned} \cos^2 x &= \frac{3}{4} \\ \cos x &= \pm\frac{\sqrt{3}}{2} \end{aligned}

Base angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$

Solutions sit in all four quadrants.

\begin{aligned} x &= \frac{\pi}{6} \quad \text{(Q1)} \\ x &= \pi - \frac{\pi}{6} \\ &= \frac{5\pi}{6} \quad \text{(Q2)} \\ x &= \pi + \frac{\pi}{6} \\ &= \frac{7\pi}{6} \quad \text{(Q3)} \\ x &= 2\pi - \frac{\pi}{6} \\ &= \frac{11\pi}{6} \quad \text{(Q4)} \end{aligned}