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Math  /  Trigonometry

Question6. 2cos⁡θ+3=02 \cos \theta+\sqrt{3}=0 −π2≤θ≤3π2-\frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}

Studdy Solution

STEP 1

1. The equation 2cos⁡θ+3=0 2 \cos \theta + \sqrt{3} = 0 is a trigonometric equation.
2. The solution for θ \theta must be within the interval −π2≤θ≤3π2-\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}.

STEP 2

1. Isolate the cosine function.
2. Solve for θ\theta using the inverse cosine function.
3. Determine all possible solutions for θ\theta within the given interval.

STEP 3

First, isolate the cosine function by subtracting 3\sqrt{3} from both sides of the equation:
2cos⁡θ+3=0 2 \cos \theta + \sqrt{3} = 0 2cos⁡θ=−3 2 \cos \theta = -\sqrt{3}

STEP 4

Now, divide both sides by 2 to solve for cos⁡θ\cos \theta:
cos⁡θ=−32 \cos \theta = -\frac{\sqrt{3}}{2}

STEP 5

Use the inverse cosine function to find the principal value of θ\theta. We know that cos⁡θ=−32\cos \theta = -\frac{\sqrt{3}}{2} corresponds to angles in the second and third quadrants:
The reference angle for cos⁡θ=32\cos \theta = \frac{\sqrt{3}}{2} is π6\frac{\pi}{6}.
Thus, the angles where cos⁡θ=−32\cos \theta = -\frac{\sqrt{3}}{2} are:
θ=π−π6 \theta = \pi - \frac{\pi}{6} θ=π+π6 \theta = \pi + \frac{\pi}{6}

STEP 6

Calculate the specific angles:
θ=π−π6=5π6 \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} θ=π+π6=7π6 \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}

STEP 7

Verify that the solutions are within the interval −π2≤θ≤3π2-\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}:
Both 5π6\frac{5\pi}{6} and 7π6\frac{7\pi}{6} fall within the specified interval.
The values of θ\theta are:
5π6,7π6 \boxed{\frac{5\pi}{6}, \frac{7\pi}{6}}

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