Solve 2cos2x−3sin x=0 for 0∘leq xleq360^∘.

Quadratic function and equation
asked 2021-01-17

Solve \(\displaystyle{2}{\cos{{2}}}{x}−{3}{\sin{{x}}}={0}{f}{\quad\text{or}\quad}{0}∘\leq{x}\leq{360}^∘.\)

Answers (1)


So that we have only one type of trig function in this, let's use the fact that
to convert that cos^2 into an expression in sin^2.
Now here's the trick: let y=sin(x). Then we have the quadratic equation
Using the quadratic formula, we can easily see that the two solutions for this are
Now substitute y=sin(x) back in to get​
The first of those has no solutions. The sine function is always between -1 and 1. So we just need the solutions to the second equation. Looking at our unit circle, we see that they are

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