question 41 solve sin(2x)+cos(x)=0 on the interval 0,2π). list exact radian answers if possible. if the…

Explanation:

Step1: Use double - angle formula

Recall $\sin(2x)=2\sin(x)\cos(x)$. The equation $\sin(2x)+\cos(x) = 0$ becomes $2\sin(x)\cos(x)+\cos(x)=0$.

Step2: Factor out $\cos(x)$

Factor the left - hand side: $\cos(x)(2\sin(x)+1)=0$.

Step3: Set each factor equal to zero

Set $\cos(x)=0$ and $2\sin(x)+1 = 0$.

• For $\cos(x)=0$, in the interval $[0,2\pi)$, $x=\frac{\pi}{2}$ or $x = \frac{3\pi}{2}$.
• For $2\sin(x)+1=0$, we have $\sin(x)=-\frac{1}{2}$. In the interval $[0,2\pi)$, $x=\frac{7\pi}{6}$ or $x=\frac{11\pi}{6}$.

Answer:

$x=\frac{\pi}{2},\frac{3\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$