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Math  /  Trigonometry

Questionsin⁡2θ+2cos⁡2θ−1=cos⁡2θ\sin ^{2} \theta+2 \cos ^{2} \theta-1=\cos ^{2} \theta

Studdy Solution

STEP 1

1. We are given a trigonometric equation involving sin⁡2θ\sin^2 \theta and cos⁡2θ\cos^2 \theta.
2. We need to simplify and solve the equation for θ\theta.
3. We can use trigonometric identities to simplify the equation.

STEP 2

1. Use trigonometric identities to simplify the equation.
2. Solve the simplified equation for θ\theta.

STEP 3

Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to express sin⁡2θ\sin^2 \theta in terms of cos⁡2θ\cos^2 \theta.
sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta
Substitute this into the original equation:
(1−cos⁡2θ)+2cos⁡2θ−1=cos⁡2θ(1 - \cos^2 \theta) + 2 \cos^2 \theta - 1 = \cos^2 \theta

STEP 4

Simplify the equation by combining like terms:
1−cos⁡2θ+2cos⁡2θ−1=cos⁡2θ1 - \cos^2 \theta + 2 \cos^2 \theta - 1 = \cos^2 \theta
This simplifies to:
cos⁡2θ=cos⁡2θ\cos^2 \theta = \cos^2 \theta

STEP 5

Since the equation cos⁡2θ=cos⁡2θ\cos^2 \theta = \cos^2 \theta is an identity, it holds true for all values of θ\theta.
Therefore, the solution to the equation is:
θ∈R\theta \in \mathbb{R}

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