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

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Answer 1 · 2016-12-21T10:35:42

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

$sin theta=0.39 :.theta=sin^-1 (0.39)=22.95^0(2dp)$ . 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]

How do you solve sintheta=0.39sintheta=0.39 in the interval 0<=theta<2pi0<=theta<2pi?