Question

If \[ 1+\cos2\theta+\cos4\theta+\cos6\theta=0, \qquad 0\le\theta\le180^\circ, \] then

• A. \(\theta=30^\circ,150^\circ,75^\circ\)
• B. \(\theta=45^\circ,135^\circ,25^\circ\)
• C. \(\theta=30^\circ,135^\circ,120^\circ\)
• D. \(\theta=30^\circ,45^\circ,90^\circ,135^\circ,150^\circ\)

Hint:

For sums of cosines: \[ \cos A+\cos B = 2\cos\frac{A+B}{2} \cos\frac{A-B}{2} \] This identity often converts complicated trigonometric equations into simple factors.

Answer 1

The Correct Option is D

Concept: Use the identity \[ 1+\cos2\theta = 2\cos^2\theta \] and \[ \cos4\theta+\cos6\theta = 2\cos5\theta\cos\theta \]

Step 1: Simplify the equation. \[\begin{aligned} 1+\cos2\theta+\cos4\theta+\cos6\theta = 0 \end{aligned}\] \[\begin{aligned} 2\cos^2\theta + 2\cos5\theta\cos\theta = 0 \end{aligned}\] \[\begin{aligned} 2\cos\theta (\cos\theta+\cos5\theta) = 0 \end{aligned}\]

Step 2: Solve the factors. Either \[ \cos\theta=0 \] giving \[ \theta=90^\circ \] or \[ \cos\theta+\cos5\theta=0 \] Using \[ \cos A+\cos B = 2\cos\frac{A+B}{2} \cos\frac{A-B}{2} \] \[\begin{aligned} 2\cos3\theta\cos2\theta=0 \end{aligned}\] Thus, \[ \cos3\theta=0 \] or \[ \cos2\theta=0 \]

Step 3: Find all solutions in the interval. From \[ \cos3\theta=0 \] \[ 3\theta=90^\circ,270^\circ,450^\circ \] \[ \theta=30^\circ,90^\circ,150^\circ \] From \[ \cos2\theta=0 \] \[ 2\theta=90^\circ,270^\circ \] \[ \theta=45^\circ,135^\circ \] Combining distinct values, \[ 30^\circ,\; 45^\circ,\; 90^\circ,\; 135^\circ,\; 150^\circ \] \[\begin{aligned} \boxed{ \theta= 30^\circ,\; 45^\circ,\; 90^\circ,\; 135^\circ,\; 150^\circ } \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.