Question

Please Help! will give brainlest!

Answer 1

Answer:

$\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

Step-by-step explanation:

we want to solve the following trigonometric equation:

$\displaystyle - 2 { \sin}^{2} (\theta )+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

The first step of solving trigonometric equation is to rewrite the equation in terms of one trigonometric function . With Pythagorean theorem, we know that sin²x=1-cos²x . It will be helpful to rewrite the equation in terms of one trig functions. Therefore, substitute 1-cos²$\theta$ in the place of sin²$\theta$:

$\displaystyle - 2 (1 - \cos ^{2} ) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

simplify:

$\displaystyle\implies - 2 (1 - \cos ^{2} (\theta) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies- 2 + \cos ^{2} (\theta)+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies\cos ^{2} (\theta)+ \cos( \theta) -1 = 0$

Consider cos² $\theta\implies$ x. Thus,

$2 {x}^{2} + x - 1 = 0$

solving the quadratic equation yields:

${x}^{} = \frac{1}{2} \\ x = - 1$

back-substitute:

$\begin{cases} \cos( \theta) = \dfrac{1}{2} \\ \cos( \theta) = - 1 \end{cases} \theta \in[0,2\pi)$

take inverse trig in both sides

$\implies \begin{cases} \theta = \dfrac{\pi}{3} + 2n\pi\\\theta= \frac{5\pi}{3} + 2n\pi \\ \theta = \pi + 2n\pi\end{cases} \theta \in[0,2\pi) \\\\\implies\theta= \frac{\pi}{3}+\dfrac{2n\pi}{3},\theta\in [0,2\pi)\\\text{when n=0}\\ \implies \theta = \frac{\pi}{3} \\ \text{when n=1}\\ \theta= \pi\\ \text{when n=2}\\\theta=\frac{5\pi}{ 3}$

In conclusion,

$\displaystyle\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$