Question

The sum of all values of $x$ in $[0, 2\pi]$, for which $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$, is equal to :

• A. $8\pi$
• B. $9\pi$
• C. $11\pi$
• D. $12\pi$

Hint:

When grouping terms, aim to produce a common factor. For symmetric sequences like $\sin x, \sin 2x, ......., \sin nx$, grouping the first with the last usually yields a common sine factor.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To solve an equation with multiple trigonometric terms, we use grouping and sum-to-product identities to factorize the expression. Then, each factor is set to zero to find the roots in the given interval.
Step 2: Key Formula or Approach:
\[ \sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right) \] Step 3: Detailed Explanation:
The equation is \(\sin x + \sin 2x + \sin 3x + \sin 4x = 0\).
Group the first and last terms, and the middle two terms:
\( (\sin 4x + \sin x) + (\sin 3x + \sin 2x) = 0 \)
Apply the sum-to-product identity:
\( 2 \sin(5x/2) \cos(3x/2) + 2 \sin(5x/2) \cos(x/2) = 0 \)
Factor out \(2 \sin(5x/2)\):
\( 2 \sin(5x/2) [ \cos(3x/2) + \cos(x/2) ] = 0 \)
Apply \(\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})\) to the bracket:
\( 2 \sin(5x/2) [ 2 \cos(x) \cos(x/2) ] = 0 \)
Now, find the values of \(x\) in \([0, 2\pi]\) for which each factor is zero:
Case 1: \(\sin(5x/2) = 0 \Rightarrow 5x/2 = n\pi \Rightarrow x = 2n\pi/5\).
Roots: \(0, 2\pi/5, 4\pi/5, 6\pi/5, 8\pi/5, 2\pi\). (Sum \(S_1 = 6\pi\))
Case 2: \(\cos(x) = 0 \Rightarrow x = \pi/2, 3\pi/2\). (Sum \(S_2 = 2\pi\))
Case 3: \(\cos(x/2) = 0 \Rightarrow x/2 = \pi/2 \Rightarrow x = \pi\). (Sum \(S_3 = \pi\))
Total sum = \(6\pi + 2\pi + \pi = 9\pi\).
Step 4: Final Answer:
The sum of all values of \(x\) is \(9\pi\).