How do you step through solving #4cos(4theta) = -2sin(theta)#?

Let #s = sin theta#. Then

#s = - 2 cos 4theta = -2 ( 1 - 2 sin^2(2theta) )#

#= - 2 ( 1 - 8 sin^2theta cos^2theta ) = - 2 ( 1 - 8 s^2 ( 1 - s^2 ))#

giving the biquadratic

#16 s^4 - 16 s^2 + s + 2 = 0#.

Graph locates s near # - 1, - 0.25, 0.5 and 1#
graph{16 x^4-16 x^2+x+2-y=0[-2 2 -1 1]}
5-sd s near #-1# is #-0.96460, giving

#theta = - 74. 71^o in [ - pi/2, pi/2 ] #, from the general solution

#( 180 k + ( - 1 )^k( - 74.71)^o, k = 0, +_ 1, +- 2, +- 3, ...#
graph{16 x^4-16 x^2+x+2-y=0[-0.9646 -0.964595-0.0001 0.0001]}

To be continued, in my 2nd answer, due to heavy graphics load, on my low-memory computer.

Graph for cross check, for #s = s_1 = -0.96460#, giving solutions

#theta = ...-105.29^o, -74.71^o, 254.71^o, ...#

#= ... - 1.838 rad, -1.304 rad, 4.445 rad, ..#

Graph for common values, from the sine waves

#sin theta and -2 cos 4theta#:
graph{(y-sin x)(y+2 cos (4x))(x+1.838+0.0001y)(x+1.304+0.0001y)(x-4.445+0.0001y)=0[-2 6 -2 2]}

Note the common points, on these #theta#-solution lines.

The second #s = -0.34258#
graph{16 x^4-16 x^2+x+2-y=0[-0.342583 -0.34258 -0.0001 0.0001]}
Correspondingly,

#theta = kpi + ( - 1 )^k arcsin (-0.34258) = ... , -159.57^o,#

#- 20.03^o, 200.03^o ... #

#= ...-2.875 rad, -0.3496 rad, 3.491 rad, ...#

To be continued.

The third 5-sd #s = 0.46243#.

Correspondingly,

#x = kpi + ( - 1 )^k arcsin ( 0.46343), k = 0, +- 1, +-2, +_ 3, ...#
graph{y-16(x^4-x^2)-x-2=0[0.43242 0.43243 -0.0001 0.0001]}
As the #s^3#-term is absent in the biquadratic,

the sum of the four s values is 0.

So, the fourth s = - ( sum of the other three ) = 0.85475.

Correspondingly,

#x = kpi + ( - 1 )^k arcsin ( 0.85475), k = 0, +- 1, +-2, +_ 3, ...#

Answer:

#x = kpi + ( - 1 )^k arcsin ( s_i), i = 1, 2, 3, 4#

#and k = 0, +- 1, +-2, +- 3, ...#, where

#{s_i}, i = 1, 2, 3, 4 = ( -0.96460 -0.34258 0.46343 0.85435 )#