+0  
 
0
69
1
avatar+9 

ll solutions in the interval 0° ≤ 𝜃 < 360°. If 2 cos 𝜃 + sin 2𝜃 = 0

 Dec 7, 2020
 #1
avatar+9192 
0

\(2\cos\theta+\sin(2\theta)=0\)

 

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°

 Dec 7, 2020

75 Online Users

avatar
avatar
avatar