Question

The sum of all the values of \(\theta\in(0,2\pi)\) satisfying the equation \[ \sin\theta+3\cos2\theta+\sin3\theta = \cos\theta+3\sin2\theta+\cos3\theta \] is

• A. \(\dfrac{5\pi}{8}\)
• B. \(\dfrac{13\pi}{8}\)
• C. \(\dfrac{32\pi}{8}\)
• D. \(\dfrac{28\pi}{8}\)

Hint:

When equations contain \(\sin\theta,\cos\theta,\sin3\theta,\cos3\theta\), convert them into sum-to-product forms. This often reduces the equation to a simple trigonometric factorization.

Answer 1

The Correct Option is D

Concept: Use the identities \[ \sin A-\cos A = \sqrt2\sin\left(A-\frac{\pi}{4}\right) \] and \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}. \] Then factor the equation and solve in \((0,2\pi)\).

Step 1: Bring all terms to one side. \[ \sin\theta-\cos\theta + 3(\cos2\theta-\sin2\theta) + (\sin3\theta-\cos3\theta) =0. \] Using \[ \sin x-\cos x = \sqrt2\sin\left(x-\frac{\pi}{4}\right), \] we get \[ \sqrt2\sin\left(\theta-\frac{\pi}{4}\right) + 3\sqrt2\cos\left(2\theta+\frac{\pi}{4}\right) + \sqrt2\sin\left(3\theta-\frac{\pi}{4}\right) =0. \] Dividing by \(\sqrt2\), \[ \sin\left(\theta-\frac{\pi}{4}\right) + \sin\left(3\theta-\frac{\pi}{4}\right) + 3\cos\left(2\theta+\frac{\pi}{4}\right) =0. \]

Step 2: Combine the sine terms. Using \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}, \] \[ \sin\left(\theta-\frac{\pi}{4}\right) + \sin\left(3\theta-\frac{\pi}{4}\right) = 2\sin\left(2\theta-\frac{\pi}{4}\right)\cos\theta. \] Hence, \[ 2\sin\left(2\theta-\frac{\pi}{4}\right)\cos\theta + 3\cos\left(2\theta+\frac{\pi}{4}\right) =0. \] Using \[ \sin\left(2\theta-\frac{\pi}{4}\right) = \frac{\sin2\theta-\cos2\theta}{\sqrt2}, \] \[ \cos\left(2\theta+\frac{\pi}{4}\right) = \frac{\cos2\theta-\sin2\theta}{\sqrt2}, \] we obtain \[ (\sin2\theta-\cos2\theta)(2\cos\theta-3)=0. \]

Step 3: Solve the factors. Since \[ 2\cos\theta-3=0 \] has no solution, \[ \sin2\theta-\cos2\theta=0. \] Therefore, \[ \tan2\theta=1. \] \[ 2\theta=\frac{\pi}{4}+n\pi. \] \[ \theta=\frac{\pi}{8}+\frac{n\pi}{2}. \]

Step 4: Find all solutions in \((0,2\pi)\). The solutions are \[ \frac{\pi}{8}, \quad \frac{5\pi}{8}, \quad \frac{9\pi}{8}, \quad \frac{13\pi}{8}. \] Their sum is \[ \frac{\pi+5\pi+9\pi+13\pi}{8} = \frac{28\pi}{8}. \] Therefore, \[ \boxed{\frac{28\pi}{8}} \] \[ \boxed{\text{Answer = (D)}} \]