4sin2(2x)+7sin2x−2=0
We have a double-angle identity that gives us:
sin(2x)=2sinxcosx then:
sin2(2x)=4sin2xcos2x
Let's plug this in:
16sin2xcos2x+7sin2x−2=0
Now, from the identity sin2x+cos2x=1, we can get:
cos2x=1−sin2x
Let's plug this in:
16sin2x(1−sin2x)+7sin2x−2=0
16sin2x−16sin4x+7sin2x−2=0
−16sin4x+23sin2x−2=0. Let's multiply the equation by −1:
16sin4x−23sin2x+2=0
Let's let z=sin2x. Then z2=sin4x. Let's plug them in:
16z2−23z+2=0
Let's use the quadratic formula to solve for z:
z=23±√529−4(16)(2)2(16)=23±√40132
z=sin2x=1.3445 and z=sin2x=0.0930
sinx=±√1.3445=±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=±√0.0930=±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
