What is the general solution for cos^2(3x)=cos(9x^2) ?

The general solution for cos^2(3x)=cos(9x^2) is x=(pi+4pin)/6 ,x=(3pi+4pin)/6 ,x=(2pin)/3

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cos^2(3x)=cos(9x^2)

cos^{2} (3 x) = cos (9 x^{2})

x = \frac{π + 4 πn}{6}, x = \frac{3 π + 4 πn}{6}, x = \frac{2 πn}{3}

Solution steps

cos^{2} (3 x) = cos (9 x^{2})

Subtract

cos (9 x^{2})

from both sides

cos^{2} (3 x) - cos (9 x^{2}) = 0

Let:

u = 3 x

cos^{2} (u) - cos (\frac{u ^{2}}{u}) = 0

Factor

cos^{2} (u) - cos (\frac{u ^{2}}{u}) : cos (u) (cos (u) - 1)

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

Solving each part separately

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

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

cos (u) - 1 = 0 : u = 2 πn

Combine all the solutions

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

Substitute back

u = 3 x

3 x = \frac{π}{2} + 2 πn : x = \frac{π + 4 πn}{6}

3 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π + 4 πn}{6}

3 x = 2 πn : x = \frac{2 πn}{3}

x = \frac{π + 4 πn}{6}, x = \frac{3 π + 4 πn}{6}, x = \frac{2 πn}{3}

Since the equation is undefined for:

\frac{π + 4 πn}{6}, \frac{3 π + 4 πn}{6}, \frac{2 πn}{3}

x = \frac{π + 4 πn}{6}, x = \frac{3 π + 4 πn}{6}, x = \frac{2 πn}{3}

2 cos (θ) - 2 = 0

x, y = r (sec (x))

csc^{2} (x) - 3 = 6 tan (x)

C, arctan (\frac{\frac{1}{2 \cdot π \cdot R \cdot C}}{f}) = 0

sin (x) = - 2 cos (x)

What is the general solution for cos^2(3x)=cos(9x^2) ?
The general solution for cos^2(3x)=cos(9x^2) is x=(pi+4pin)/6 ,x=(3pi+4pin)/6 ,x=(2pin)/3