Trigonometry

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

2 cos^2 theta - 5 cos theta + 10 = 13      subtract 13 from both sides....and we can write

2 [cos θ]^2  - 5 cos θ - 3   = 0        now...just factor with θ  as we would with "x"

[ 2 cos θ  + 1]  [cos θ  - 3 ]    = 0

Set each factor to 0

2 cos θ  + 1  = 0    →     cos θ  =  -1/2       and this happens at  2pi/3 + n (2pi)     and at  4pi/3 + n(2pi)    where n is any integer  

For the other factor  we have that   cos  θ  - 3 = 0   →   cos θ  = 3      and this is impossible

See??...it wasn't really that bad, was it ???

Thanks for the reply CPhill, so would the final answer by theta= 2pi/3 and theta= 4pi/3 ???

2 cos^2 theta - 5 cos theta + 10 = 13

Thanks for the reply CPhill, so would the final answer by theta= 2pi/3 and theta= 4pi/3 ???

Yep...if we just wanted the values between  0  and 2pi