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Question Number 117387 by otchereabdullai@gmail.com last updated on 11/Oct/20
Solvethetrigonometricequation5sinθ+3=0forvalueofθfrom0°to360°
Commented by bemath last updated on 11/Oct/20
θ=sin−1(−35)+k.360°θ=180°−sin−1(35)+k.360°wheresin−1(−35)=−0.643501
Answered by TANMAY PANACEA last updated on 11/Oct/20
sinθ=−35=−0.6=−sin37osinθ=sin(180o+37o)=sin217osinθ=sin(360o−37o)=sin323o
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