sin(4x)-cos(2x)=0

Solution

sin (4 x) - cos (2 x) = 0

x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}, x = \frac{π + 12 πn}{12}, x = \frac{5 π + 12 πn}{12}

Solution steps

sin (4 x) - cos (2 x) = 0

Let:

u = 2 x

sin (2 u) - cos (u) = 0

Rewrite using trig identities

- cos (u) + 2 cos (u) sin (u) = 0

Factor

- cos (u) + 2 cos (u) sin (u) : cos (u) (2 sin (u) - 1)

cos (u) (2 sin (u) - 1) = 0

Solving each part separately

cos (u) = 0 or 2 sin (u) - 1 = 0

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

2 sin (u) - 1 = 0 : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Combine all the solutions

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Substitute back

u = 2 x

2 x = \frac{π}{2} + 2 πn : x = \frac{π + 4 πn}{4}

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

2 x = \frac{π}{6} + 2 πn : x = \frac{π + 12 πn}{12}

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π + 12 πn}{12}

x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}, x = \frac{π + 12 πn}{12}, x = \frac{5 π + 12 πn}{12}

\frac{6}{π} = 3 cos (2 x)

sin (2 x) = - 2 cos (x)

4 - 8 cos^{2} (x) = 0

cos^{2} (x) = \frac{1 - cos ( 2 x )}{2}

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