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Rearrange the given equation to isolate the sine and cosine terms.

Mathematics

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Answer

θ≈36.10∘\theta \approx 36.10^\circ

Step-by-step solution

7 steps

  1. Rearrange the given equation to isolate the sine and cosine terms.

    9sin⁡2θ−28cos⁡2θ=09 \sin 2\theta - 28 \cos 2\theta = 0 9sin⁡2θ=28cos⁡2θ9 \sin 2\theta = 28 \cos 2\theta

  2. Divide both sides by cos⁡2θ\cos 2\theta (assuming cos⁡2θ≠0\cos 2\theta \neq 0) to express the equation in terms of tan⁡2θ\tan 2\theta.

    9sin⁡2θcos⁡2θ=28cos⁡2θcos⁡2θ\frac{9 \sin 2\theta}{\cos 2\theta} = \frac{28 \cos 2\theta}{\cos 2\theta} 9tan⁡2θ=289 \tan 2\theta = 28

  3. Solve for tan⁡2θ\tan 2\theta.

    tan⁡2θ=289\tan 2\theta = \frac{28}{9} tan⁡2θ≈3.1111\tan 2\theta \approx 3.1111

  4. Find the principal value for 2θ2\theta by taking the inverse tangent.

    2θ=arctan⁡(289)2\theta = \arctan\left(\frac{28}{9}\right) 2θ≈72.19∘2\theta \approx 72.19^\circ

  5. Determine the range for 2θ2\theta.

    Given 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ, multiply by 2 to find the range for 2θ2\theta.

    0∘≤2θ≤180∘0^\circ \le 2\theta \le 180^\circ

  6. Since tan⁡2θ\tan 2\theta is positive, 2θ2\theta must be in the first quadrant.

    The value found in Step 4 is the only solution within the range 0∘≤2θ≤180∘0^\circ \le 2\theta \le 180^\circ.

    2θ≈72.19∘2\theta \approx 72.19^\circ

  7. Solve for θ\theta.

    θ=72.19∘2\theta = \frac{72.19^\circ}{2} θ≈36.095∘\theta \approx 36.095^\circ

    Rounding to two decimal places:

    θ≈36.10∘\theta \approx 36.10^\circ
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