+9 ll solutions in the interval 0° ≤ 𝜃 < 360°. If 2 cos 𝜃 + sin 2𝜃 = 0
+9479 \(2\cos\theta+\sin(2\theta)=0\)
Let's use the double-angle formula for sin which says \(\sin(2\theta)=2\sin\theta\cos\theta\)
\(2\cos\theta+2\sin\theta\cos\theta=0\)
Now we can factor 2 cos θ out of both terms on the left side of the equation.
\(2\cos\theta(1+\sin\theta)=0\) (Notice if we distribute 2 cos θ we get the previous expression)
Set each factor equal to 0 and solve for θ:
\(\begin{array}{ccc} 2\cos\theta=0&\text{or}&1+\sin\theta=0\\~\\ \cos\theta=0&\text{or}&\sin\theta=-1\\~\\ \theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\dots&\text{or}&\theta=\frac{3\pi}{2},\frac{7\pi}{2},\frac{11\pi}{2},\frac{15\pi}{2},\dots\\~\\ \theta=90^\circ,270^\circ,450^\circ,630^\circ,\dots&\text{or}&\theta=270^\circ,630^\circ,990^\circ,1350^\circ\dots \end{array}\)
The solutions in the interval [0°, 360°) are: 90°, 270°
