Question

How do you solve `sintheta=0.39` in the interval `0<=theta<2pi`?

Answer 1

In interval `0<=x<2pi, theta=22.95^0 ,theta=157.05^0`

Explanation:

`sin theta=0.39 :.theta=sin^-1 (0.39)=22.95^0(2dp)`. Also `sin(pi-22.95)=0.39 :.theta=180-22.95=157.05^0`

Solution: In interval `0<=x<2pi, theta=22.95^0 ,theta=157.05^0`[Ans]