Question

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

Answer 1

`\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)`