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Math  /  Trigonometry

QuestionSolve the equation on the interval 0≤x≤2π0 \leq x \leq 2 \pi a. 4sin⁡θ+33=34 \sin \theta+3 \sqrt{3}=\sqrt{3}

Studdy Solution

STEP 1

1. The equation 4sin⁡θ+33=3 4 \sin \theta + 3 \sqrt{3} = \sqrt{3} is a trigonometric equation.
2. We will solve for θ\theta within the interval 0≤θ≤2π0 \leq \theta \leq 2\pi.

STEP 2

1. Isolate the trigonometric function sin⁡θ\sin \theta.
2. Solve for θ\theta using inverse trigonometric functions.
3. Determine the solutions within the given interval.

STEP 3

First, isolate sin⁡θ\sin \theta by subtracting 333 \sqrt{3} from both sides of the equation:
4sin⁡θ+33=3 4 \sin \theta + 3 \sqrt{3} = \sqrt{3} 4sin⁡θ=3−33 4 \sin \theta = \sqrt{3} - 3 \sqrt{3} 4sin⁡θ=−23 4 \sin \theta = -2 \sqrt{3}

STEP 4

Divide both sides by 4 to solve for sin⁡θ\sin \theta:
sin⁡θ=−234 \sin \theta = \frac{-2 \sqrt{3}}{4} sin⁡θ=−32 \sin \theta = -\frac{\sqrt{3}}{2}

STEP 5

Determine the values of θ\theta where sin⁡θ=−32\sin \theta = -\frac{\sqrt{3}}{2} within the interval 0≤θ≤2π0 \leq \theta \leq 2\pi.
The sine function is −32-\frac{\sqrt{3}}{2} at angles θ=4π3\theta = \frac{4\pi}{3} and θ=5π3\theta = \frac{5\pi}{3}.
The solutions for θ\theta in the interval 0≤θ≤2π0 \leq \theta \leq 2\pi are:
θ=4π3,5π3 \theta = \frac{4\pi}{3}, \frac{5\pi}{3}

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