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Solve. 2sin x cos x=sin x

Solve.\newline2sin⁡xcos⁡x=sin⁡x 2 \sin x \cos x=\sin x

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Full solution

Q. Solve.\newline2sin⁡xcos⁡x=sin⁡x 2 \sin x \cos x=\sin x
  1. Write Equation: First, let's write down the given equation.\newline2sin⁡xcos⁡x=sin⁡x2\sin x \cos x = \sin x
  2. Subtract and Rearrange: Now, we subtract sin⁡x\sin x from both sides to move all terms involving xx to one side of the equation.\newline2sin⁡xcos⁡x−sin⁡x=02\sin x \cos x - \sin x = 0
  3. Factor Out: Factor out sin⁡x\sin x from the left side of the equation.sin⁡x(2cos⁡x−1)=0\sin x (2\cos x - 1) = 0
  4. Set Equal to Zero: Set each factor equal to zero to find the solutions for xx.sin⁡x=0\sin x = 0 and 2cos⁡x−1=02\cos x - 1 = 0
  5. Solve sin⁡x=0\sin x = 0: Solve the first equation sin⁡x=0\sin x = 0. The solutions for xx are the angles where the sine function equals zero. x=nπx = n\pi, where nn is an integer.
  6. Solve cos⁡x=12\cos x = \frac{1}{2}: Solve the second equation 2cos⁡x−1=02\cos x - 1 = 0.\newlineAdd 11 to both sides and then divide by 22.\newlinecos⁡x=12\cos x = \frac{1}{2}
  7. Solve cos⁡x=12\cos x = \frac{1}{2}: Solve the second equation 2cos⁡x−1=02\cos x - 1 = 0. Add 11 to both sides and then divide by 22. cos⁡x=12\cos x = \frac{1}{2}Find the values of xx where the cosine function equals 12\frac{1}{2}. x=π3+2nπx = \frac{\pi}{3} + 2n\pi or x=5π3+2nπx = \frac{5\pi}{3} + 2n\pi, where nn is an integer.

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