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

The general solution for [sin(4x)cos(x)-cos(4x)sin(x)]^2=1 is x= pi/2+(2pin)/3 ,x= pi/6+(2pin)/3

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[sin(4x)cos(x)-cos(4x)sin(x)]^2=1

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

x = \frac{π}{2} + \frac{2 πn}{3}, x = \frac{π}{6} + \frac{2 πn}{3}

Solution steps

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

Subtract

1

from both sides

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

Factor

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

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

Solving each part separately

sin (4 x) cos (x) - cos (4 x) sin (x) + 1 = 0 or sin (4 x) cos (x) - cos (4 x) sin (x) - 1 = 0

sin (4 x) cos (x) - cos (4 x) sin (x) + 1 = 0 : x = \frac{π}{2} + \frac{2 πn}{3}

sin (4 x) cos (x) - cos (4 x) sin (x) - 1 = 0 : x = \frac{π}{6} + \frac{2 πn}{3}

Combine all the solutions

x = \frac{π}{2} + \frac{2 πn}{3}, x = \frac{π}{6} + \frac{2 πn}{3}

cos (θ) = \frac{202}{331 \cdot 2 33}

5 sin (θ) + 12 cos (θ) = 13

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

cot (x) = \frac{5}{12}, π < x < \frac{3 π}{2}

tan (x) = \frac{1}{5}, tan (x + \frac{π}{2})

What is the general solution for [sin(4x)cos(x)-cos(4x)sin(x)]^2=1 ?
The general solution for [sin(4x)cos(x)-cos(4x)sin(x)]^2=1 is x= pi/2+(2pin)/3 ,x= pi/6+(2pin)/3