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Convert the equation to a single trigonometric function using the identity 2 theta = 1 − 2 theta.

Computer Science

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Answer

θ=0∘,180∘,360∘\theta = 0^\circ, 180^\circ, 360^\circ

Step-by-step solution

5 steps

here's how to solve the equation 2cos⁡2θ−3sin⁡θ−2=02\cos^2 \theta - 3\sin \theta - 2 = 0 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ:

  1. Convert the equation to a single trigonometric function using the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta.

    2(1−sin⁡2θ)−3sin⁡θ−2=02(1 - \sin^2 \theta) - 3\sin \theta - 2 = 0
  2. Expand and simplify the equation into a quadratic form in terms of sin⁡θ\sin \theta.

    2−2sin⁡2θ−3sin⁡θ−2=02 - 2\sin^2 \theta - 3\sin \theta - 2 = 0 −2sin⁡2θ−3sin⁡θ=0-2\sin^2 \theta - 3\sin \theta = 0

    Multiply by −1-1 to make the leading coefficient positive:

    2sin⁡2θ+3sin⁡θ=02\sin^2 \theta + 3\sin \theta = 0
  3. Factor out sin⁡θ\sin \theta.

    sin⁡θ(2sin⁡θ+3)=0\sin \theta (2\sin \theta + 3) = 0
  4. Set each factor equal to zero and solve for sin⁡θ\sin \theta.

    Case 1: sin⁡θ=0\sin \theta = 0
    Case 2: 2sin⁡θ+3=0  ⟹  2sin⁡θ=−3  ⟹  sin⁡θ=−322\sin \theta + 3 = 0 \implies 2\sin \theta = -3 \implies \sin \theta = -\frac{3}{2}

  5. Find the values of θ\theta for each valid solution within the range 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.

    For Case 1: sin⁡θ=0\sin \theta = 0
    The angles where sin⁡θ=0\sin \theta = 0 in the given range are 0∘0^\circ, 180∘180^\circ, and 360∘360^\circ.

    For Case 2: sin⁡θ=−32\sin \theta = -\frac{3}{2}
    Since the range of the sine function is [−1,1][-1, 1], sin⁡θ=−32\sin \theta = -\frac{3}{2} (which is −1.5-1.5) has no real solutions.

    The solutions for θ\theta are:

    θ=0∘,180∘,360∘\theta = 0^\circ, 180^\circ, 360^\circ
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