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Question Number 117387 by otchereabdullai@gmail.com last updated on 11/Oct/20

Solve the trigonometric equation  5sinθ+3=0 for value of θ from 0° to  360°

Solvethetrigonometricequation5sinθ+3=0forvalueofθfrom0°to360°

Commented by bemath last updated on 11/Oct/20

θ = sin^(−1)  (−(3/5))+k.360°  θ = 180°−sin^(−1) ((3/5))+k.360°  where sin^(−1) (−(3/5))= −0.643501

θ=sin1(35)+k.360°θ=180°sin1(35)+k.360°wheresin1(35)=0.643501

Answered by TANMAY PANACEA last updated on 11/Oct/20

sinθ=((−3)/5)=−0.6=−sin37^o   sinθ=sin(180^o +37^o )=sin217^o   sinθ=sin(360^o −37^o )=sin323^o

sinθ=35=0.6=sin37osinθ=sin(180o+37o)=sin217osinθ=sin(360o37o)=sin323o

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