Trigonometry: Question 9

Worked solution

Part (a): Rewriting the equation

Start from the identity connecting $\sin^2 x$ and $\cos^2 x$: $\sin^2 x + \cos^2 x \equiv 1 \quad\Longrightarrow\quad \cos^2 x \equiv 1 - \sin^2 x$

Substitute this into the left-hand side of $2\cos^2 x - 5\sin x = 4$: $2(1-\sin^2 x) - 5\sin x = 4$

Expand the bracket: $2 - 2\sin^2 x - 5\sin x = 4$

Collect every term on one side (subtract $4$ from both sides): $-2\sin^2 x - 5\sin x + 2 - 4 = 0$

$-2\sin^2 x - 5\sin x - 2 = 0$

Multiply through by $-1$ so that the $\sin^2 x$ term is positive: $2\sin^2 x + 5\sin x + 2 = 0$

as required.

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

Let $s = \sin x$, so the equation becomes: $2s^2 + 5s + 2 = 0$

Factorise: $(2s+1)(s+2) = 0$

(Check: $(2s+1)(s+2) = 2s^2+4s+s+2 = 2s^2+5s+2$ ✓.)

So either: $2s+1=0 \Rightarrow s=\sin x = -\frac12 \qquad\text{or}\qquad s+2=0 \Rightarrow s=\sin x = -2$

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

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

The reference angle is $\dfrac{\pi}{6}$, since $\sin\dfrac{\pi}{6}=\tfrac12$. Sine is negative in the third and fourth quadrants, so: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \qquad\text{and}\qquad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

Both lie within $0 \le x \le 2\pi$.

Check (substituting into the original equation): at $x=\tfrac{7\pi}{6}$, $\sin x=-\tfrac12$ and $\cos x=-\tfrac{\sqrt3}{2}$, so $\cos^2 x=\tfrac34$, giving $2\left(\tfrac34\right)-5\left(-\tfrac12\right)=\tfrac32+\tfrac52=4$ ✓. At $x=\tfrac{11\pi}{6}$, $\sin x=-\tfrac12$ and $\cos x=\tfrac{\sqrt3}{2}$, so $\cos^2 x=\tfrac34$ again, giving the same value $4$ ✓.

Final answers

• (a) $2\sin^2 x + 5\sin x + 2 = 0$, as shown above.
• (b) $x = \boxed{\dfrac{7\pi}{6},\ \dfrac{11\pi}{6}}$