Question

Solve the equation for 0 < x < 2pi (2sin x - 1)(2cos^2 x - 1) = 0

Answer 1

A product is zero if and only if at least one of the two factors is zero. So, you need either

$2\sin(x)-1=0$

or

$2\cos^2(x)-1=0$

In the first case, the equation reduces to

$2\sin(x) - 1 \iff 2\sin(x)=1\iff \sin(x)=\dfrac{1}{2}$

In the second case, we have

$2\cos^2(x)-1=0 \iff 2\cos^2(x)=1 \iff \cos^2(x)=\dfrac{1}{2} \iff \cos(x)=\pm\dfrac{1}{\sqrt{2}}$

So, you need one of the following three to happen:

$\sin(x)=\dfrac{1}{2},\quad \cos(x)=-\dfrac{1}{\sqrt{2}},\quad \cos(x)=\dfrac{1}{\sqrt{2}}$

I assume you have a table to look up for these known values. If you have trouble solving for $x$, feel free to hit me up.