Question

How do you find the exact value of `2cos^2theta+3sintheta-2cos2theta=1` in the interval `0<=theta<2pi`?

Answer 1

`sin^(-1)(1/3)= 19.47^o and (180-19,47)^o=160.53^o`, nearly.

Explanation:

`2(1-sin^2theta)+3sin theta-2(1-2sin^2theta)=3 sin theta =1`. So,

#sin theta = sin (sin^(-1)(1/3)1/3. So, the general solution is

#theta = kpi+(-1)^ksin(-1)(1/3), k = 9, +-1, +-2, ..

For k = 0 1nd 1, we have the answer