SOLUTION: Find all solutions in the interval [0,2pi) 2cos^2x-3=3sinx

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Question 1178312: Find all solutions in the interval [0,2pi)
2cos^2x-3=3sinx
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

2cos%5E2%28x%29-3=3sin%28x%29
2cos%5E2%28x%29=3sin%28x%29%2B3
cos%282+x%29+%2B+1=3sin%28x%29%2B3
cos%282+x%29+=3sin%28x%29%2B3-1
cos%5E2%28x%29+-+sin%5E2%28x%29+=3sin%28x%29%2B2............cos%5E2%28x%29=1-sin%5E2%28x%29+
1-sin%5E2%28x%29++-+sin%5E2%28x%29+=3sin%28x%29%2B2
1-2sin%5E2%28x%29+=3sin%28x%29%2B2
2sin%5E2%28x%29%2B3sin%28x%29%2B2-1=0
2sin%5E2%28x%29%2B3sin%28x%29%2B1=0
2sin%5E2%28x%29%2B2sin%28x%29%2Bsin%28x%29%2B1=0
%282sin%5E2%28x%29%2Bsin%28x%29%29%2B%282sin%28x%29%2B1%29=0
sin%28x%29%282sin%28x%29%2B1%29%2B%282sin%28x%29%2B1%29=0
%28sin%28x%29%2B1%29%282sin%28x%29%2B1%29=0
solutions:
if %28sin%28x%29%2B1%29=0=>sin%28x%29=-1
then
x=sin%5E-1%28-1%29
x=+-pi%2F2 ....periodicity of sin is 2pi
x=+2pi-pi%2F2
x=+3pi%2F2%2B2pi%2An
in interval [0,2pi)
x=+270°

if 2sin%28x%29%2B1=0->sin%28x%29=-1%2F2+
x=sin%5E-1%28-1%2F2%29
x=-pi%2F6
x=2pi-pi%2F6
x=11pi%2F6%2B2pi%2An
x=+7pi%2F6%2B2pi%2An
in interval [0,2pi)
+x=+210°
x=+330°