Question

2cos^2 = 3sin solve for every angle in a circle

Answer 1

If I understand this correctly, the problem statement is \[2\cos^2 \theta = 3\sin \theta\]On a unit circle\[\cos^2 \theta +\sin^2 \theta =x^2+y^2=1 \implies 2x^2=3y ~and~x^2=1-y^2 \implies 2-2y^2=3y\]\[\implies 2y^2+3y-2=0\]That should be enough to solve for y. From that you can get the corresponding values of x, which will in turn give you values of theta.

Answer 2

Hope that was helpful.

Answer 3

don't we ave to get it down to one thing? like cosine or sign ?

Answer 4

Remember that on the unit circle x = cos theta and y = sin theta. After you have the values of x and y, take the arccos and arcsin to get theta. I think you will probably have four solutions.

Answer 5

\[2y^2+3y-2=0 \implies y^2+\frac{3}{2}y=1\implies y^2+\frac{3}{2}y+\frac{9}{4}=1+\frac{9}{4}\]\[\implies (y+\frac{3}{2})^2=\frac{13}{4}\implies y+\frac{3}{2}=\pm \frac{\sqrt{13}}{2}\implies \sin \theta = y=\frac{-2 \pm \sqrt{13}}{2}\]Note that there is only one value of y that works. If you let the plus or minus be minus, it gives a value for sin theta with an absolute value a lot bigger than one. The solution for y matches with two corresponding values of x to yield two solutions.