Question

Use inverse functions where needed to find all solutions of the equation in the interval 2 cos2 x + 9 sin x = 6

Answer 1

You can use the identity $\cos(2x) = \cos^2(x)-\sin^2(x)$ to write the equation as

$2(\cos^2(x)-\sin^2(x))+ 9\sin(x) = 6$

Now, from the fundamental equation of trigonometry, deduce an expression for $\cos^2(x)$ in terms of $\sin^2(x)$:

$\cos^2(x)+\sin^2(x) = 1 \implies \cos^2(x) = 1-\sin^2(x)$

The equation becomes

$2(1-\sin^2(x)-\sin^2(x))+ 9\sin(x) = 6 \iff 2(1-2\sin^2(x))+ 9\sin(x) = 6$

Simplify the left hand side and move all terms to left hand side:

$2-4\sin^2(x)+ 9\sin(x) -6 = 0 \iff -4\sin^2(x) + 9\sin(x) - 4 = 0$

Now, if you let $t = \sin(x)$, this equation becomes a quadratic equation:

$-4t^2 + 9t - 4 = 0$

The two solutions of this equations are

$t = \cfrac{9}{8} - \cfrac{\sqrt{17}}{8},\quad t = \cfrac{9}{8} + \cfrac{\sqrt{17}}{8}$

We must be careful, because we have to remember that $t$ was actually $\sin(x)$. This means that $t$ can only assume values between -1 and 1. The second solution exceeds 1, so we reject it. So, we have

$t = \cfrac{9}{8} - \cfrac{\sqrt{17}}{8} \implies \sin(x) = \cfrac{9}{8} - \cfrac{\sqrt{17}}{8} \implies x = \arcsin\left(\cfrac{9}{8} - \cfrac{\sqrt{17}}{8}\right)$