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Answer to Question #10629 in Trigonometry for nhbn

Question #10629
sin^2*2θ+6sin2θ+5=0 i have to solve for x
Expert's answer
Let's denote sin(2θ)=t. Note that |t|<=1
So we obtain
t^2+6t+5=0 - quadratic equation.
Let's solve it for t:
t1=-1, t2=-5
T2 doesn't fit us because |-5|=5>1. So we have single solution
therefore sin(2θ)=-1
2θ=-pie/2+2 pie k where k-integer
whence θ=-pie/4+pie k - answer

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