Solve for x 1-cos(12x)=2sin(6x)^2

$1 - \cos (12 x) = 2 \sin^{2} (6 x)$

Move $2 \sin^{2} (6 x)$ to the left side of the equation by subtracting it from both sides.

$1 - \cos (12 x) - 2 \sin^{2} (6 x) = 0$

Simplify the left side.

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Move $- 2 \sin^{2} (6 x)$ .

$1 - 2 \sin^{2} (6 x) - \cos (12 x) = 0$

Apply the cosine double–angle identity.

$\cos (2 (6 x)) - \cos (12 x) = 0$

Simplify by adding terms.

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Multiply $6$ by $2$ .

$\cos (12 x) - \cos (12 x) = 0$

Subtract $\cos (12 x)$ from $\cos (12 x)$ .

$0 = 0$

Since $0 = 0$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x 1-cos(12x)=2sin(6x)^2

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