5.10 Solving trigonometric equations in an interval — Practise with help

1. Both equal $\dfrac{1}{2}$.
2. $x^2 - x - 2 = (x - 2)(x + 1)$.
3. $x = 2$ or $x = -1$.

If question 2 was hard, revise factorising a quadratic before you continue.

Sine is positive in the first and second quadrants. One solution is $\dfrac{\pi}{3}$.

The second solution is $\pi - \dfrac{\pi}{3}$.

Cosine is negative in the second and third quadrants. The related acute angle is $\dfrac{\pi}{3}$.

The two angles are $\pi - \dfrac{\pi}{3}$ and $\pi + \dfrac{\pi}{3}$.

Tangent has period $\pi$. The principal value is $\dfrac{\pi}{3}$.

Add $\pi$ to the principal value to get the second solution, then check it is still below $2\pi$.

Let $x = 2\theta$. The new interval is $0 \le x < 4\pi$, which is two full turns.

In $0 \le x < 4\pi$, sine takes $\dfrac{\sqrt{3}}{2}$ at $x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{7\pi}{3}, \dfrac{8\pi}{3}$. Divide each by $2$.

Factorise the left-hand side. Do not divide by $\cos\theta$.

$\cos\theta(2\cos\theta - 1) = 0$, so $\cos\theta = 0$ or $\cos\theta = \dfrac{1}{2}$.

For $\cos\theta = 0$:

For $\cos\theta = \dfrac{1}{2}$:

The solutions are $\dfrac{\pi}{3}$, $\dfrac{\pi}{2}$, $\dfrac{3\pi}{2}$ and $\dfrac{5\pi}{3}$.

$\sin\theta = \dfrac{1}{2}$ or $\sin\theta = -\dfrac{1}{2}$. Each gives two solutions.

For $\sin\theta = \dfrac{1}{2}$, the angles are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$. For $\sin\theta = -\dfrac{1}{2}$, they are $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$.