Solve-the-trigonometric-equation-5sin-3-0-for-value-of-from-0-to-360-

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

Solve the trigonometric equation

5 \sin θ + 3 = 0 for value of θ from 0 ° to

360 °

Commented by bemath last updated on 11/Oct/20

θ = \sin^{- 1} (- \frac{3}{5}) + k . 360 °

θ = 180 ° - \sin^{- 1} (\frac{3}{5}) + k . 360 °

where \sin^{- 1} (- \frac{3}{5}) = - 0.643501

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

s i n θ = \frac{- 3}{5} = - 0.6 = - s i n 37^{o}

s i n θ = s i n (180^{o} + 37^{o}) = s i n 217^{o}

s i n θ = s i n (360^{o} - 37^{o}) = s i n 323^{o}