Question

solve the equation in degrees for all exact solutions where appropriate. round approximate answers in degrees to the nearest tenth. write answers using the least possible nonnegative angle measures.
4 sin θ = 4 cos 2θ
what is the solution set?
a. {0° + 180°n, 120° + 180°n, 240° + 180°n, where n is any integer}
b. {30°, + 360°n, 120° + 360°n, 270° + 360°n, where n is any integer}
c. {0° + 360°n, 150° + 360°n, 240° + 360°n, where n is any integer}
d. {30° + 360°n, 150° + 360°n, 270° + 360°n, where n is any integer}

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula \(\cos2\theta = 1 - 2\sin^{2}\theta\).
The given equation \(4\sin\theta=4\cos2\theta\) becomes \(4\sin\theta = 4(1 - 2\sin^{2}\theta)\).
Divide both sides by \(4\): \(\sin\theta=1 - 2\sin^{2}\theta\).
Rearrange to get a quadratic equation: \(2\sin^{2}\theta+\sin\theta - 1 = 0\).
Let \(x = \sin\theta\), then the equation is \(2x^{2}+x - 1=0\).

Step2: Solve the quadratic equation

Factor the quadratic equation \(2x^{2}+x - 1=(2x - 1)(x + 1)=0\).
Set each factor equal to zero:

• For \(2x - 1 = 0\), we have \(x=\sin\theta=\frac{1}{2}\).

When \(\sin\theta=\frac{1}{2}\), \(\theta = 30^{\circ}+360^{\circ}n\) or \(\theta = 150^{\circ}+360^{\circ}n\), \(n\in\mathbb{Z}\) (using the unit - circle, \(\sin\theta=y/r\), and \(y = \frac{1}{2}\) in the first and second quadrants).

• For \(x + 1=0\), we have \(x=\sin\theta=-1\).

When \(\sin\theta=-1\), \(\theta = 270^{\circ}+360^{\circ}n\), \(n\in\mathbb{Z}\) (using the unit - circle, \(y=-1\) in the fourth quadrant).

Answer:

D. \(\{30^{\circ}+360^{\circ}n,150^{\circ}+360^{\circ}n,270^{\circ}+360^{\circ}n,\text{where }n\text{ is any integer}\}\)