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Solve for θ if − 6 sin θ 4 = 3 √ 3 4 and 0 ≤ θ < 2 π .

Sagot :

Answer:

Step-by-step explanation:

Hey there! To solve the equation \( -6 \sin(4\theta) = 3 \sqrt{34} \) for \( 0 \leq \theta < 2\pi \), we need to isolate \( \sin(4\theta) \) first.

Divide both sides by -6:

\( \sin(4\theta) = -\frac{3 \sqrt{34}}{6} = -\frac{\sqrt{34}}{2} \)

Now, to find the values of \( \theta \) that satisfy this equation, we need to look at the unit circle and find where the sine function equals \( -\frac{\sqrt{34}}{2} \).

The reference angle for \( \sin^{-1}(\frac{\sqrt{34}}{2}) \) is \( \frac{\pi}{6} \), which corresponds to \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \) in the second and third quadrants, respectively.

Since we are looking for \( 0 \leq \theta < 2\pi \), the solutions are \( \frac{7\pi}{6} \) and \( \frac{11\pi}{6} \).

So, the answers are \( \theta = \frac{7\pi}{6} \) and \( \theta = \frac{11\pi}{6} \).