I have to find the value of $\theta$ in this equation "$\cos2\theta \sin\theta = 1$"
The problem is that I have to solve using algebra (making equations…etc) like this one, for example: \begin{align} 4\sin^2\theta – 1 = 0\\ \sin^2\theta = 0.25\\ \sin\theta = +0.5, -0.5\\ \theta = 30, 150, 210, 330\end{align}
But for "$\cos2\theta \sin\theta = 1$", I figured, according to the fact that $\cos\alpha$ can be only between -1 and 1. And the same for $\sin\alpha$, that the only two expected solutions are when $\cos2\theta$ and $\sin\theta$ are both equal to 1 or -1 so that the product is equal to 1 as in the main equation.
So $\theta = 270$. \begin{align} \cos(540)\sin(270) = 1\\ \cos(180)\sin(270) = 1\\ -1 * -1 = 1 \end{align}
I'm trying to find another way to solve it like the way I solved the other example $4\sin^2\theta – 1 = 0$..
Best Answer
You need formulae for double angle, e.g. in here: $ \cos (2 \theta) = 1 - 2 (\sin (\theta))^2 $
Then let $x = \sin (\theta)$ to get
$$ 1 = \cos (2 \theta) \sin (\theta) = (1 - 2 (\sin (\theta))^2) \sin (\theta) = (1-2x^2)x $$
Solving that third order equation gives as the only real root $x = -1$, hence $-1 = \sin (\theta)$ or
$$ \theta = \frac32 \pi $$
and of course you may add multiples of $2 \pi$.