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Math  /  Trigonometry

Questionc. cos⁡2θ−sin⁡2θ+sin⁡θ=0\cos ^{2} \theta-\sin ^{2} \theta+\sin \theta=0

Studdy Solution

STEP 1

1. We are solving the trigonometric equation cos⁡2θ−sin⁡2θ+sin⁡θ=0\cos^2 \theta - \sin^2 \theta + \sin \theta = 0.
2. The equation involves trigonometric identities that might simplify the expression.
3. Solutions will be found for θ\theta in terms of standard trigonometric angles.

STEP 2

1. Use trigonometric identities to simplify the equation.
2. Rearrange the equation to facilitate solving for θ\theta.
3. Solve for θ\theta using algebraic methods.
4. Verify solutions within the context of trigonometric functions.

STEP 3

Use the trigonometric identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta to simplify the equation:
cos⁡2θ−sin⁡2θ+sin⁡θ=0 \cos^2 \theta - \sin^2 \theta + \sin \theta = 0
Substitute cos⁡2θ\cos^2 \theta with 1−sin⁡2θ1 - \sin^2 \theta:
(1−sin⁡2θ)−sin⁡2θ+sin⁡θ=0 (1 - \sin^2 \theta) - \sin^2 \theta + \sin \theta = 0

STEP 4

Simplify the equation by combining like terms:
1−2sin⁡2θ+sin⁡θ=0 1 - 2\sin^2 \theta + \sin \theta = 0

STEP 5

Rearrange the equation to form a quadratic in terms of sin⁡θ\sin \theta:
−2sin⁡2θ+sin⁡θ+1=0 -2\sin^2 \theta + \sin \theta + 1 = 0
Multiply the entire equation by −1-1 to make it easier to solve:
2sin⁡2θ−sin⁡θ−1=0 2\sin^2 \theta - \sin \theta - 1 = 0

STEP 6

Solve the quadratic equation 2sin⁡2θ−sin⁡θ−1=02\sin^2 \theta - \sin \theta - 1 = 0 using the quadratic formula:
The quadratic formula is given by sin⁡θ=−b±b2−4ac2a\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=2a = 2, b=−1b = -1, and c=−1c = -1.
Calculate the discriminant:
b2−4ac=(−1)2−4(2)(−1)=1+8=9 b^2 - 4ac = (-1)^2 - 4(2)(-1) = 1 + 8 = 9

STEP 7

Substitute the values into the quadratic formula:
sin⁡θ=−(−1)±92(2) \sin \theta = \frac{-(-1) \pm \sqrt{9}}{2(2)}
sin⁡θ=1±34 \sin \theta = \frac{1 \pm 3}{4}
This gives two solutions:
sin⁡θ=44=1 \sin \theta = \frac{4}{4} = 1
sin⁡θ=−24=−12 \sin \theta = \frac{-2}{4} = -\frac{1}{2}

STEP 8

Find θ\theta for each solution:
1. sin⁡θ=1\sin \theta = 1 implies θ=π2+2kπ\theta = \frac{\pi}{2} + 2k\pi, where kk is an integer.
2. sin⁡θ=−12\sin \theta = -\frac{1}{2} implies θ=7π6+2kπ\theta = \frac{7\pi}{6} + 2k\pi or θ=11π6+2kπ\theta = \frac{11\pi}{6} + 2k\pi, where kk is an integer.
The solutions for θ\theta are θ=π2+2kπ\theta = \frac{\pi}{2} + 2k\pi, θ=7π6+2kπ\theta = \frac{7\pi}{6} + 2k\pi, and θ=11π6+2kπ\theta = \frac{11\pi}{6} + 2k\pi, where kk is an integer.

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