What is the general solution for sin(8x)-2cos(4x)=0 ?

The general solution for sin(8x)-2cos(4x)=0 is x=(pi+4pin)/8 ,x=(3pi+4pin)/8

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sin(8x)-2cos(4x)=0

sin (8 x) - 2 cos (4 x) = 0

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

Solution steps

sin (8 x) - 2 cos (4 x) = 0

Let:

u = 4 x

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

Substitute back

u = 4 x

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

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

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

1.812 = 316 \cdot cos (1.496 \cdot x) + 1.496

t, cos (wt + d) = 0

sin (θ) = 3 cos (θ), 0 \leq θ < 2 π

cos (2 x) - \frac{1}{3} cos (x) = 0

3 = 23 sin (x)

What is the general solution for sin(8x)-2cos(4x)=0 ?
The general solution for sin(8x)-2cos(4x)=0 is x=(pi+4pin)/8 ,x=(3pi+4pin)/8