4sin^2(2x)+7sin^2x-2=0
We have a double-angle identity that gives us:
sin(2x)=2sinxcosx then:
sin^2(2x)=4sin^2xcos^2x
Let's plug this in:
16sin^2xcos^2x+7sin^2x-2=0
Now, from the identity sin^2x+cos^2x=1, we can get:
cos^2x=1-sin^2x
Let's plug this in:
16sin^2x(1-sin^2x)+7sin^2x-2=0
16sin^2x-16sin^4x+7sin^2x-2=0
-16sin^4x+23sin^2x-2=0. Let's multiply the equation by -1:
16sin^4x-23sin^2x+2=0
Let's let z=sin^2x. Then z^2=sin^4x. Let's plug them in:
16z^2-23z+2=0
Let's use the quadratic formula to solve for z:
z=(23+-sqrt(529-4(16)(2)))/(2(16))=(23+-sqrt401)/32
z=sin^2x=1.3445 and z=sin^2x=0.0930
sinx=+-sqrt1.3445=+-1.1595, then x=arcsin(+-1.1595)
These two answers are not acceptable because we are looking for angles whose sin values are larger than +-1.
sinx=+-sqrt0.0930=+-0.3050, then x=arcsin(+-0.3050)
x=17.76 Degrees, and x=-17.76=360-17.76=342.24Degrees
Now, because the problem asked for values between 0 and 360 Degrees, we can have two more answers:
x=180-17.76=162.24 Degrees, and
x=180+17.76=197.76 Degrees