Question

5sin^2theta-4sintheta+cos2theta=0

Answer 1

Remove cos^2 theta using 1-sin^2 theta, then\[5\sin^2 \theta - 4 \sin \theta +1 - \sin ^ 2 \theta =0\]\[=4\sin^2 \theta - 4 \sin \theta +1 =0\]The quadratic equation then gives,\[\sin \theta = \frac{1}{2}\](there's a double root).

The general solution is then\[\theta = n.180^o+(-1)^n \sin ^{-1}\frac{1}{2}\]i.e.\[\theta = n.180^o+(-1)^n 30^o\]

Answer 2

for n any integer.

Answer 3

so what would the answer be in radians?

Answer 4

Well 180 degrees is equivalent to pi radians, and 30 degrees is (30/180)pi = pi/6 radians.

Answer 5

\[\theta = n \pi +(-1)^n \frac{\pi}{6}\]

Answer 6

You're welcome.