chapter 9: review score: 110/280 answered: 11/28 question 21 find all solutions of the equation 2 sin²x…

Explanation:

Step1: Use trigonometric identity

Recall $\sin^{2}x = 1-\cos^{2}x$. Substitute it into the equation $2\sin^{2}x-\cos x = 1$. We get $2(1 - \cos^{2}x)-\cos x=1$.

Step2: Rearrange the equation

Expand and rearrange to get a quadratic - equation in terms of $\cos x$. $2 - 2\cos^{2}x-\cos x = 1$, which simplifies to $2\cos^{2}x+\cos x - 1=0$.

Step3: Solve the quadratic equation

Let $t = \cos x$. The quadratic equation $2t^{2}+t - 1 = 0$ can be factored as $(2t - 1)(t + 1)=0$. Set each factor equal to zero:

• If $2t - 1=0$, then $t=\cos x=\frac{1}{2}$.
• If $t + 1=0$, then $t=\cos x=-1$.

Step4: Find the values of $x$ in the given interval

When $\cos x=\frac{1}{2}$ and $x\in[0,2\pi)$, $x=\frac{\pi}{3}$ or $x = \frac{5\pi}{3}$. When $\cos x=-1$ and $x\in[0,2\pi)$, $x=\pi$.

Answer:

$x_1=\frac{\pi}{3}$, $x_2=\pi$, $x_3=\frac{5\pi}{3}$