Math  /  Trigonometry

QuestionFind all solutions of the given equation. (Enter your answers as a comma-separated list. Let kk be any integer. Do not round coefficients of kk. Other terms can be rounded to two decimal places where appropriate.) 2cos(θ)3=0θ=rad\begin{array}{r} 2 \cos (\theta)-\sqrt{3}=0 \\ \theta=\square \mathrm{rad} \end{array}

Studdy Solution
Since cosine is periodic with a period of 2π2\pi, the general solutions for θ\theta are:
θ=π6+2kπandθ=11π6+2kπ \theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2k\pi
where kk is any integer.
The solutions for θ\theta are:
θ=π6+2kπ,11π6+2kπ \theta = \frac{\pi}{6} + 2k\pi, \frac{11\pi}{6} + 2k\pi

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