Multiple choice

The $\theta_{1},\theta_{2},\theta_{3},\theta_{4}$ are the roots of the equation $\sin (\theta+\alpha)=k \sin 2 \theta$, no two of which differ by multiple of $2\pi$, then $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}$ is equal to

  1. $2 n \pi, n \epsilon Z$
  2. $(2n+1)\pi, n \epsilon Z$
  3. $n \pi, n \epsilon Z$
  4. $None\ of\ these$
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A Correct answer
Explanation

The equation sin(θ + α) = k sin(2θ) can be expanded using trigonometric identities to form a polynomial in sin(θ) or cos(θ). The sum of roots for such trigonometric equations often relates to the periodicity of the functions involved.

AI explanation

The given equation is sin(theta + alpha) = k sin(2theta). Expanding the left side and replacing sin(2theta) with 2sin(theta)cos(theta) gives sin(theta)cos(alpha) + cos(theta)sin(alpha) - 2ksin(theta)cos(theta) = 0. Factoring out sin(theta) yields sin(theta)(cos(alpha) + sin(alpha)cot(theta) - 2kcos(theta)) = 0, showing that either sin(theta) = 0 or cos(alpha) + sin(alpha)cot(theta) - 2kcos(theta) = 0. The equation sin(theta) = 0 provides two roots whose sum is an even multiple of pi, like 0 and pi. For the quadratic in cot(theta), the sum of its two roots is 2k/sin(alpha), which corresponds to the sum of the cotangents of the two remaining angles, pi plus an angle and 2pi minus that angle, meaning their sum is also an even multiple of pi. Adding these together shows the total sum of all four roots is 2npi.

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