# 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

t=-1

therefore sin(2θ)=-1

2θ=-pie/2+2 pie k where k-integer

whence θ=-pie/4+pie k - answer

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

t=-1

therefore sin(2θ)=-1

2θ=-pie/2+2 pie k where k-integer

whence θ=-pie/4+pie k - answer

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