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Solve the trigonometric equation: (2theta - 30^) - ((2theta)/(3)) = 0

Mathematics

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Question 12:

  1. Rewrite the equation.

    The given equation is:

    sin⁡(2θ−30∘)−cos⁡(2θ3)=0\sin(2\theta - 30^\circ) - \cos\left(\frac{2\theta}{3}\right) = 0

    Rearrange the equation:

    sin⁡(2θ−30∘)=cos⁡(2θ3)\sin(2\theta - 30^\circ) = \cos\left(\frac{2\theta}{3}\right)
  2. Use the trigonometric identity cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

    Apply this identity to the right side of the equation:

    cos⁡(2θ3)=sin⁡(90∘−2θ3)\cos\left(\frac{2\theta}{3}\right) = \sin\left(90^\circ - \frac{2\theta}{3}\right)

    Substitute this back into the equation:

    sin⁡(2θ−30∘)=sin⁡(90∘−2θ3)\sin(2\theta - 30^\circ) = \sin\left(90^\circ - \frac{2\theta}{3}\right)
  3. Solve for θ\theta using the general solutions for sin⁡A=sin⁡B\sin A = \sin B.

    The general solutions for sin⁡A=sin⁡B\sin A = \sin B are A=B+360∘nA = B + 360^\circ n or A=(180∘−B)+360∘nA = (180^\circ - B) + 360^\circ n, where nn is an integer.

    Case 1: 2θ−30∘=90∘−2θ3+360∘n2\theta - 30^\circ = 90^\circ - \frac{2\theta}{3} + 360^\circ n
    Combine terms with θ\theta:

    2θ+2θ3=90∘+30∘+360∘n2\theta + \frac{2\theta}{3} = 90^\circ + 30^\circ + 360^\circ n 6θ+2θ3=120∘+360∘n\frac{6\theta + 2\theta}{3} = 120^\circ + 360^\circ n 8θ3=120∘+360∘n\frac{8\theta}{3} = 120^\circ + 360^\circ n

    Multiply by 3:

    8θ=3(120∘+360∘n)8\theta = 3(120^\circ + 360^\circ n) 8θ=360∘+1080∘n8\theta = 360^\circ + 1080^\circ n

    Divide by 8:

    θ=360∘8+1080∘8n\theta = \frac{360^\circ}{8} + \frac{1080^\circ}{8} n θ=45∘+135∘n\theta = 45^\circ + 135^\circ n

    For n=0n=0, θ=45∘\theta = 45^\circ.
    For n=1n=1, θ=45∘+135∘=180∘\theta = 45^\circ + 135^\circ = 180^\circ.
    For n=2n=2, θ=45∘+270∘=315∘\theta = 45^\circ + 270^\circ = 315^\circ.

    Case 2: 2θ−30∘=(180∘−(90∘−2θ3))+360∘n2\theta - 30^\circ = (180^\circ - (90^\circ - \frac{2\theta}{3})) + 360^\circ n

    2θ−30∘=180∘−90∘+2θ3+360∘n2\theta - 30^\circ = 180^\circ - 90^\circ + \frac{2\theta}{3} + 360^\circ n 2θ−30∘=90∘+2θ3+360∘n2\theta - 30^\circ = 90^\circ + \frac{2\theta}{3} + 360^\circ n

    Combine terms with θ\theta:

    2θ−2θ3=90∘+30∘+360∘n2\theta - \frac{2\theta}{3} = 90^\circ + 30^\circ + 360^\circ n 6θ−2θ3=120∘+360∘n\frac{6\theta - 2\theta}{3} = 120^\circ + 360^\circ n 4θ3=120∘+360∘n\frac{4\theta}{3} = 120^\circ + 360^\circ n

    Multiply by 3:

    4θ=3(120∘+360∘n)4\theta = 3(120^\circ + 360^\circ n) 4θ=360∘+1080∘n4\theta = 360^\circ + 1080^\circ n

    Divide by 4:

    θ=360∘4+1080∘4n\theta = \frac{360^\circ}{4} + \frac{1080^\circ}{4} n θ=90∘+270∘n\theta = 90^\circ + 270^\circ n

    For n=0n=0, θ=90∘\theta = 90^\circ.

    Assuming the solutions are required in the range 0∘≤θ<360∘0^\circ \le \theta < 360^\circ, the values for θ\theta are:
    $\boxed{\theta = 45^\circ, 90^\circ, 180^\circ, 315

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