solve the following equation graphically over the interval 0, 2π). 2 sin x = 1 - 2 cos x\nthe solution set…

solve the following equation graphically over the interval 0, 2π). 2 sin x = 1 - 2 cos x\nthe solution set over the interval 0, 2π) is {}. (type an integer or decimal rounded to the nearest hundredth as needed. use a comma to separate answers as needed.)

solve the following equation graphically over the interval 0, 2π). 2 sin x = 1 - 2 cos x\nthe solution set over the interval 0, 2π) is {}. (type an integer or decimal rounded to the nearest hundredth as needed. use a comma to separate answers as needed.)

Answer

Explanation:

Step1: Rewrite the equation

Rewrite $2\sin x = 1 - 2\cos x$ as $2\sin x+2\cos x = 1$. Then, $\sin x+\cos x=\frac{1}{2}$.

Step2: Use the identity $\sin x+\cos x=\sqrt{2}\sin(x + \frac{\pi}{4})$

We get $\sqrt{2}\sin(x+\frac{\pi}{4})=\frac{1}{2}$, so $\sin(x + \frac{\pi}{4})=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$.

Step3: Find the general solutions

$x+\frac{\pi}{4}=\arcsin(\frac{\sqrt{2}}{4})+2k\pi$ or $x+\frac{\pi}{4}=\pi-\arcsin(\frac{\sqrt{2}}{4})+2k\pi$, $k\in\mathbb{Z}$.

Step4: Solve for $x$ in the interval $[0, 2\pi)$

For $x+\frac{\pi}{4}=\arcsin(\frac{\sqrt{2}}{4})+2k\pi$: $x=\arcsin(\frac{\sqrt{2}}{4})-\frac{\pi}{4}+2k\pi$. When $k = 0$, $x=\arcsin(\frac{\sqrt{2}}{4})-\frac{\pi}{4}\approx0.36 - 0.79=- 0.43$ (not in the interval $[0,2\pi)$). For $x+\frac{\pi}{4}=\pi-\arcsin(\frac{\sqrt{2}}{4})+2k\pi$: $x=\pi-\arcsin(\frac{\sqrt{2}}{4})-\frac{\pi}{4}+2k\pi=\frac{3\pi}{4}-\arcsin(\frac{\sqrt{2}}{4})+2k\pi$. When $k = 0$, $x=\frac{3\pi}{4}-\arcsin(\frac{\sqrt{2}}{4})\approx2.36 - 0.36 = 2.00$. Also, considering the periodicity of the sine - function, for $x+\frac{\pi}{4}=2\pi+\arcsin(\frac{\sqrt{2}}{4})$ (from the general solution of $\sin t=\sin\alpha\Rightarrow t=\alpha + 2k\pi$ or $t=\pi-\alpha+2k\pi$), $x = 2\pi+\arcsin(\frac{\sqrt{2}}{4})-\frac{\pi}{4}\approx6.28+0.36 - 0.79 = 5.85$.

Answer:

$2.00,5.85$