Solve Trigonometric Equations

This tutorial explains how to solve trigonometric equations using identities, algebraic manipulation, and the unit circle. Each example includes clear reasoning, graphical interpretation, and complete solutions. The unit circle is especially useful for identifying reference angles and locating all possible solutions.

For additional practice, see: Grade 12 Trigonometric Equations with Detailed Solutions.


Worked Examples with Full Solutions

Example 1

Solve the trigonometric equation (find all solutions):

\[ 2\cos x + 2 = 3 \]

Solution

Start by isolating the trigonometric function.

\[ 2\cos x = 1 \quad \Rightarrow \quad \cos x = \frac{1}{2} \]

The cosine function equals \(\frac{1}{2}\) at two angles in the interval \( [0,2\pi) \):

\[ x_1 = \frac{\pi}{3}, \quad x_2 = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \] Cosine solutions on the unit circle

Since cosine has period \(2\pi\), all solutions are obtained by adding multiples of \(2\pi\):

\[ x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2k\pi, ; k \in \mathbb{Z} \]

Conclusion: The equation has infinitely many solutions.


Example 2

Find all solutions in the interval \( [0,2\pi) \):

\[ -5\cos^2 x + 9\sin x = -3 \]

Solution

Rewrite \(\cos^2 x\) using the identity \(\cos^2 x = 1 - \sin^2 x\).

\[ -5(1 - \sin^2 x) + 9\sin x = -3 \]

Simplify:

\[ 5\sin^2 x + 9\sin x - 2 = 0 \]

Let \(u = \sin x\). This transforms the equation into a quadratic.

\[ 5u^2 + 9u - 2 = 0 \]

Using the quadratic formula:

\[ u = \frac{-9 \pm \sqrt{121}}{10} \] \[ u_1 = -2 \quad (\text{invalid, since } -1 \le \sin x \le 1) \] \[ u_2 = 0.2 \]

Now solve \(\sin x = 0.2\). Sine is positive in Quadrants I and II.

Sine solutions on the unit circle \[ x_1 = \arcsin(0.2), \quad x_2 = \pi - \arcsin(0.2) \]

Conclusion: There are exactly two solutions in \( [0,2\pi)\).


Example 3

Find all solutions:

\[ \cot x\cos^2 x = \cot x \]

Solution

Subtract \(\cot x\) from both sides and factor.

\[ \cot x(\cos^2 x - 1) = 0 \]

This gives two equations:

\[ \cot x = 0 \quad \text{or} \quad \cos^2 x - 1 = 0 \]

From \(\cot x = 0\):

\[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]

From \(\cos^2 x - 1 = 0\), we get \(\cos x = \pm 1\), which corresponds to \(x = k\pi\). However, \(\cot x\) is undefined at these values.

Conclusion:

\[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]

Final Note: Solving trigonometric equations often involves the same strategies used in algebra—factoring, substitutions, and careful consideration of domain restrictions.