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Solve the given equation. (Enter your answers as a comma-separated list. Let k k be any integer.\newlinesin(θ)=12 \sin(\theta) = -\frac{1}{2}

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Q. Solve the given equation. (Enter your answers as a comma-separated list. Let k k be any integer.\newlinesin(θ)=12 \sin(\theta) = -\frac{1}{2}
  1. Identify Sine Value: The equation sin(θ)=12\sin(\theta) = -\frac{1}{2} implies that θ\theta is an angle whose sine is 12-\frac{1}{2}. We know that the sine function has a value of 12-\frac{1}{2} at specific angles in the third and fourth quadrants of the unit circle, where the sine values are negative.
  2. Find Reference Angle: The reference angle for which the sine has a value of 12\frac{1}{2} is 3030 degrees or π6\frac{\pi}{6} radians. In the third and fourth quadrants, we need to find the angles that have this reference angle but with a negative sine value.
  3. Third Quadrant Angle: In the third quadrant, the angle with a sine of 12-\frac{1}{2} is 180180 degrees + 3030 degrees, which is 210210 degrees or in radians, π+π6\pi + \frac{\pi}{6}, which simplifies to 7π6\frac{7\pi}{6}.
  4. Fourth Quadrant Angle: In the fourth quadrant, the angle with a sine of 12-\frac{1}{2} is 360360 degrees 30- 30 degrees, which is 330330 degrees or in radians, 2ππ62\pi - \frac{\pi}{6}, which simplifies to 11π6\frac{11\pi}{6}.
  5. General Solutions: Since the sine function is periodic with a period of 2π2\pi, we can add any integer multiple of 2π2\pi to our solutions to get the general form of the solutions. Therefore, the general solutions are (7π)/6+2kπ(7\pi)/6 + 2k\pi and (11π)/6+2kπ(11\pi)/6 + 2k\pi, where kk is any integer.

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