What is the general solution for cos(4x)+1=3sin(2x) ?

The general solution for cos(4x)+1=3sin(2x) is x= pi/(12)+pin,x=(5pi)/(12)+pin

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cos(4x)+1=3sin(2x)

cos (4 x) + 1 = 3 sin (2 x)

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solution steps

cos (4 x) + 1 = 3 sin (2 x)

Subtract

3 sin (2 x)

from both sides

cos (4 x) + 1 - 3 sin (2 x) = 0

Rewrite using trig identities

(1 - sin^{2} (2 x)) \cdot 2 - 3 sin (2 x) = 0

Solve by substitution

sin (2 x) = - 2, sin (2 x) = \frac{1}{2}

sin (2 x) = - 2 :

No Solution

sin (2 x) = \frac{1}{2} : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Combine all the solutions

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

tan (2 t + 15) = cot (6 t - 5)

3 sin (θ) = 0

2 cos^{3} (x) - cos^{2} (x) + 2 cos (x) - 1 = 0

x, - 44 sin (2 x) + 42 cos (2 x) = 0

sin^{2} (x) - sin (x) = 1 - sin^{2} (x)

What is the general solution for cos(4x)+1=3sin(2x) ?
The general solution for cos(4x)+1=3sin(2x) is x= pi/(12)+pin,x=(5pi)/(12)+pin