find all solutions of the equation in the interval $0,2\\pi)$.\n$2\\cos\\theta+\\sqrt{2}=0$\nwrite your…

find all solutions of the equation in the interval $0,2\\pi)$.\n$2\\cos\\theta+\\sqrt{2}=0$\nwrite your answer in radians in terms of $\\pi$,\nif there is more than one solution, separate them with commas.\n$\\theta=\\square$

find all solutions of the equation in the interval $0,2\\pi)$.\n$2\\cos\\theta+\\sqrt{2}=0$\nwrite your answer in radians in terms of $\\pi$,\nif there is more than one solution, separate them with commas.\n$\\theta=\\square$

Answer

Explanation:

Step1: Solve for $\cos\theta$

$$ \begin{align*} 2\cos\theta+\sqrt{2}& = 0\ 2\cos\theta&=-\sqrt{2}\ \cos\theta&=-\frac{\sqrt{2}}{2} \end{align*} $$

Step2: Find $\theta$ in $[0,2\pi)$

We know that $\cos\theta =-\frac{\sqrt{2}}{2}$. The cosine function is negative in the second and third quadrants. The reference angle $\theta_{ref}$ for which $\cos\theta_{ref}=\frac{\sqrt{2}}{2}$ is $\theta_{ref}=\frac{\pi}{4}$. In the second quadrant, $\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$. In the third quadrant, $\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}$.

Answer:

$\frac{3\pi}{4},\frac{5\pi}{4}$