Question

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

Answer 1

Answer:

$\theta_1=0\\ \\\theta_2=\pi\\ \\\theta_3=\dfrac{7\pi}{6}\\ \\\theta_4=\dfrac{11\pi}{6}$

Step-by-step 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:

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

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

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

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