sin^4 2(theta) - 2 sin^2 2(theta) = -1 again...like the last one, we have
[sin (2θ)]^4 - 2 [sin 2θ] + 1 = 0 now ....factor with sin2θ instead of "x"
[ ( (sin (2θ) )^2 - 1 ] [ ( (sin (2θ) )^2 - 1 ] = 0 which implies that
[ ( (sin (2θ) )^2 - 1 ] ^2 = 0 take the square root of both sides
(sin (2θ) )^2 - 1 = 0 which implies that
[sin 2θ + 1 ] [ sin2θ - 1 ] = 0 set each factor to 0
sin2θ + 1 = 0 → sin2θ = -1 since sinθ = - 1 at 3pi/2 + n(2pi) then the angle we're interested in is 1/2 of this = 3pi/4 + n (pi)
And the genereal solution to this part is θ = 3pi/4 + (2pi)n where n is an integer
The other solution is similar
sin 2θ - 1 = 0 → sin2θ = 1 since sinθ = 1 at pi/2 + n(2pi) then the angle we're interested in is 1/2 of this = pi/4 + n (pi)
And the general solution to this part is θ = pi/4 + n(pi) where n is an integer
So.......the solutions are 3pi/4 + n (pi) and pi/4 + n(pi) where n is an integer
Check the graph here : https://www.desmos.com/calculator/n4mki8w2rp
