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

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

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sin(2x)=cos(4x)

sin (2 x) = cos (4 x)

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

Solution steps

sin (2 x) = cos (4 x)

Subtract

cos (4 x)

from both sides

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

Let:

u = 2 x

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

Rewrite using trig identities

- 1 + sin (u) + 2 sin^{2} (u) = 0

Solve by substitution

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

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

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = 2 x

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}

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

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

6 cos^{2} (θ) + 7 = - 17 cos (θ)

0 = - 9 \cdot sin (2 x)

sec^{2} (x) = 4 tan^{2} (x)

5 sin (x) = - 4

- π sin (π x) = 0

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