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2sin theta cos theta+cos theta=0

2sin⁡θcos⁡θ+cos⁡θ=0 2 \sin \theta \cos \theta+\cos \theta=0

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Full solution

Q. 2sin⁡θcos⁡θ+cos⁡θ=0 2 \sin \theta \cos \theta+\cos \theta=0
  1. Factor out cos⁡(θ)\cos(\theta): Factor out cos⁡(θ)\cos(\theta) from the equation.2sin⁡(θ)cos⁡(θ)+cos⁡(θ)=cos⁡(θ)(2sin⁡(θ)+1)2\sin(\theta)\cos(\theta) + \cos(\theta) = \cos(\theta)(2\sin(\theta) + 1)
  2. Set equal to zero: Set the factored equation equal to zero to find the values of θ\theta that satisfy the equation.cos⁡(θ)(2sin⁡(θ)+1)=0\cos(\theta)(2\sin(\theta) + 1) = 0
  3. Solve for theta: Solve for theta by setting each factor equal to zero.\newlineFirst, cos⁡(θ)=0\cos(\theta) = 0.
  4. Find cos⁡(θ)\cos(\theta) values: Find the values of θ\theta where cos⁡(θ)=0\cos(\theta) = 0.θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}
  5. Set other factor to zero: Now, set the other factor equal to zero: 2sin⁡(θ)+1=02\sin(\theta) + 1 = 0.
  6. Solve for sin⁡(θ)\sin(\theta): Solve for sin⁡(θ)\sin(\theta).sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2}
  7. Find sin⁡(θ)\sin(\theta) values: Find the values of θ\theta where sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2}.\newlineθ=7π6,11π6\theta = \frac{7\pi}{6}, \frac{11\pi}{6}

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