Trigonometry: Question 5

Worked solution

Part (a): Rewriting the equation

Use the identity $\tan x \equiv \dfrac{\sin x}{\cos x}$ to rewrite the left-hand side: $2\sin x\tan x = 2\sin x \cdot \frac{\sin x}{\cos x} = \frac{2\sin^2 x}{\cos x}$

So the equation becomes: $\frac{2\sin^2 x}{\cos x} = 3$

Multiply both sides by $\cos x$ (valid since $\tan x$ is defined only where $\cos x \ne 0$): $2\sin^2 x = 3\cos x$

Now use $\sin^2 x \equiv 1 - \cos^2 x$: $2(1-\cos^2 x) = 3\cos x$

$2 - 2\cos^2 x = 3\cos x$

Collect all terms on one side: $2 - 2\cos^2 x - 3\cos x = 0$

Multiply through by $-1$ so the $\cos^2 x$ term is positive: $2\cos^2 x + 3\cos x - 2 = 0$

as required.

Part (b): Solving the quadratic in $\cos x$

Let $c = \cos x$: $2c^2 + 3c - 2 = 0$

Factorise: $(2c - 1)(c + 2) = 0$

(Check: $(2c-1)(c+2) = 2c^2 + 4c - c - 2 = 2c^2 + 3c - 2$ ✓.)

So either: $c = \cos x = \frac12 \qquad\text{or}\qquad c = \cos x = -2$

Since $-1 \le \cos x \le 1$ for every real $x$, the solution $\cos x = -2$ is impossible and must be rejected. Only $\cos x = \tfrac12$ remains.

Solving $\cos x = \dfrac12$ for $0 \le x \le 2\pi$:

The principal value is $x = \dfrac{\pi}{3}$ (since $\cos\dfrac{\pi}{3} = \tfrac12$ is the standard exact value at $60^\circ$). Cosine is also positive in the fourth quadrant, giving the second solution: $x = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3}-\frac{\pi}{3} = \frac{5\pi}{3}$

Both lie within $0 \le x \le 2\pi$, and at neither value is $\cos x = 0$, so $\tan x$ is defined throughout.

Check: at $x=\tfrac{\pi}{3}$, $\sin x = \tfrac{\sqrt3}{2}$, $\tan x=\sqrt3$, so $2\sin x\tan x = 2\cdot\tfrac{\sqrt3}{2}\cdot\sqrt3 = 3$ ✓. At $x=\tfrac{5\pi}{3}$, $\sin x=-\tfrac{\sqrt3}{2}$, $\tan x=-\sqrt3$, so $2\sin x\tan x = 2\cdot\left(-\tfrac{\sqrt3}{2}\right)\cdot(-\sqrt3)=3$ ✓.

Final answers

• (a) $2\cos^2 x + 3\cos x - 2 = 0$, as shown above.
• (b) $x = \boxed{\dfrac{\pi}{3},\ \dfrac{5\pi}{3}}$