trig equation

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Solve \(\cos^2 x + 3 \sin x - 3 = 0\) for 0 <= x <= 2*pi.

Guest Jul 26, 2020

geno3141 · Answer 1 · 2020-07-26T09:31:26

Use the identity: cos²(x) = 1 - sin²(x) so that the equation only conains the sin function.

cos²(x) + 3sin(x) - 3 = 0 ---> ( 1 - sin²(x) ) + 3sin(x) - 3 = 0

-sin²(x) + 3sin(x) - 2 = 0

sin²(x) - 3sin(x) + 2 = 0

( sin(x) - 2 )( sin(x) - 1 ) = 0

either sin(x) = 2 [impossible] or sin(x) = 1 ---> x = pi/2