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Solve sin theta + 1 = cos 2 theta on the interval 0 less than or equal to theta <2 pi

User Trevir
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1 Answer

2 votes

Answer:


\theta_1=0\\ \\\theta_2=\pi\\ \\\theta_3=(7\pi)/(6)\\ \\\theta_4=(11\pi)/(6)

Explanation:

Solve the equation


\sin \theta +1=\cos 2\theta

First, use formula


\cos 2\theta =1-2\sin^2\theta

Then the equation is


\sin\theta+1=1-2\sin^2\theta\\ \\\sin\theta=-2\sin^2\theta\\ \\\sin\theta+2\sin^2\theta=0\\ \\\sin\theta(1+2\sin\theta)=0

The produat is equal to 0 when one of its factors is 0:


\sin\theta=0\text{ or }1+2\sin\theta=0

Solve each of these equations:


\sin\theta=0\\ \\\theta=\pi k,\ k\in Z


1+2\sin\theta=0\\ \\\sin\theta=-(1)/(2)\\ \\\theta=(-1)^k\arcsin\left(-(1)/(2)\right)+\pi k,\ kin Z\\ \\\theta=(-1)^k\cdot\left(-(\pi)/(6)\right)+\pi k,\ k\in Z

The solutions, which are in interval
0\le \rheta<2\pi are


\theta_1=0\\ \\\theta_2=\pi\\ \\\theta_3=(-1)\cdot \left(-(\pi)/(6)\right)+\pi=(7\pi)/(6)\\ \\\theta_4=(-1)^2\cdot \left(-(\pi)/(6)\right)+2\pi=(11\pi)/(6)

User Koxon
by
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