Question

Solve the equation for the interval [0°, 360°):
4cos^2(theta)sin(theta)+2cos^2(theta)=0

Answer 1

Notice that

\[4\cos ^2(\theta )\sin (\theta )+2 \cos ^2(\theta )=2\cos ^2(\theta ) (2 \sin (\theta
)+1) =0


\]

Answer 2

So
\[
\cos ^2(\theta )=0 \\or\\
\sin(\theta) =-\frac 12

\]

Answer 3

\[
\cos ^2(\theta )=0\\
\theta =\frac \pi 2\\
\theta =3\frac \pi 2\\
\]

Answer 4

\[
\sin(\theta) =-\frac 12 \\
\theta=\frac { 7 \pi}6\\

\theta= \frac{11 \pi }{6}
\]

Answer 5

Do you understand?

Answer 6

It's supposed to be a set of 5 intervals, and its in radians. However, I do understand your answer.