What is the general solution for 3cos(4x)-2= 3/2 sin(8x)-2 ?

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

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3cos(4x)-2= 3/2 sin(8x)-2

3 cos (4 x) - 2 = \frac{3}{2} sin (8 x) - 2

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

Solution steps

3 cos (4 x) - 2 = \frac{3}{2} sin (8 x) - 2

Subtract

\frac{3}{2} sin (8 x) - 2

from both sides

3 cos (4 x) - \frac{3}{2} sin (8 x) = 0

Simplify

3 cos (4 x) - \frac{3}{2} sin (8 x) : \frac{6 cos ( 4 x ) - 3 sin ( 8 x )}{2}

\frac{6 cos ( 4 x ) - 3 sin ( 8 x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

6 cos (4 x) - 3 sin (8 x) = 0

Let:

u = 4 x

6 cos (u) - 3 sin (2 u) = 0

Rewrite using trig identities

6 cos (u) - 6 cos (u) sin (u) = 0

Factor

6 cos (u) - 6 cos (u) sin (u) : - 6 cos (u) (sin (u) - 1)

- 6 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}

4 \cdot cos (x) sin (x) + 3 \cdot cos (x) = 0

(cot (x)) (tan (x) - 1) = 0

10 tan (θ) sec (θ) = 10 cot (θ) csc (θ)

arctan (x) = \frac{( 2 \cdot 1.8 )}{[ ( 3170 + 1.7 ) - 1 ]}

cos (4 x) + sin (\frac{x}{2}) = 0

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