#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=342.24#Degrees
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
