SOLUTION: solve cos2x-2sinx-cos^2 x = -3 on the interval x ∈ [0,2π]

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Question 1192743: solve cos2x-2sinx-cos^2 x = -3 on the interval x ∈ [0,2π]
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

cos%282x%29-2sin%28x%29-cos%5E2+%28x%29+=+-3 on the interval x ∈ [0,2%CF%80]

since cos%282x%29=cos%5E2%28x%29+-+sin%5E2%28x%29, we have

cos%5E2%28x%29+-+sin%5E2%28x%29-2sin%28x%29-cos%5E2+%28x%29+=+-3+
-sin%5E2%28x%29-2sin%28x%29=+-3+
-3sin%5E2%28x%29=+-3
+sin%5E2%28x%29=+-3%2F-3+
+sin%5E2%28x%29=+1
+sin%28x%29=+sqrt%281%29
sin%28x%29=+1

x=sin%5E-1%281%29
x=pi%2F2

periodicity of sin is 2pi
general solution:
x=pi%2F2%2B2pi%2An

on the interval x ∈ [0,2%CF%80] : x=pi%2F2