What is the general solution for (1+cos(4x))sin(4x)=2cos^2(2x) ?

The general solution for (1+cos(4x))sin(4x)=2cos^2(2x) is x= pi/4+pin,x=(3pi)/4+pin,x= pi/8+(pin)/2

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(1+cos(4x))sin(4x)=2cos^2(2x)

(1 + cos (4 x)) sin (4 x) = 2 cos^{2} (2 x)

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn, x = \frac{π}{8} + \frac{πn}{2}

Solution steps

(1 + cos (4 x)) sin (4 x) = 2 cos^{2} (2 x)

Subtract

2 cos^{2} (2 x)

from both sides

sin (4 x) (1 + cos (4 x)) - 2 cos^{2} (2 x) = 0

Rewrite using trig identities

- 2 cos^{2} (2 x) + 2 cos^{2} (2 x) sin (4 x) = 0

Factor

- 2 cos^{2} (2 x) + 2 cos^{2} (2 x) sin (4 x) : 2 cos^{2} (2 x) (sin (4 x) - 1)

2 cos^{2} (2 x) (sin (4 x) - 1) = 0

Solving each part separately

cos^{2} (2 x) = 0 or sin (4 x) - 1 = 0

cos^{2} (2 x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

sin (4 x) - 1 = 0 : x = \frac{π}{8} + \frac{πn}{2}

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn, x = \frac{π}{8} + \frac{πn}{2}

\frac{5 sin ( x ) + 6}{sin ( x )} = 17

tan^{2} (x) sec (x) - 1 = 0

sin^{2} (x) + cos^{4} (x) = 2

(1 + sin^{2} (x)) = (4 cos (x))^{2}

cos (x) = 53

What is the general solution for (1+cos(4x))sin(4x)=2cos^2(2x) ?
The general solution for (1+cos(4x))sin(4x)=2cos^2(2x) is x= pi/4+pin,x=(3pi)/4+pin,x= pi/8+(pin)/2