Questions: Show that the equation (3 cos θ-tan θ) cos θ=2 can be written as a sin^2 θ+b sin θ+c=0 where a>0 and a, b, and c are integers to be found.

Show that the equation (3 cos θ-tan θ) cos θ=2 can be written as a sin^2 θ+b sin θ+c=0 where a>0 and a, b, and c are integers to be found.
Transcript text: Show that the equation \[ (3 \cos \theta-\tan \theta) \cos \theta=2 \] can be written as \[ a \sin ^{2} \theta+b \sin \theta+c=0 \] where $a>0$ and $a, b$ and $c$ are integers to be found.
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Solution

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Solution Steps

To transform the given trigonometric equation into the desired quadratic form, we need to use trigonometric identities and algebraic manipulation. Specifically, we will:

  1. Express tan⁡θ\tan \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta.
  2. Substitute tan⁡θ\tan \theta and simplify the equation.
  3. Rearrange the equation to match the form asin⁡2θ+bsin⁡θ+c=0a \sin^2 \theta + b \sin \theta + c = 0.
Step 1: Substitute tan⁡θ\tan \theta with sin⁡θcos⁡θ\frac{\sin \theta}{\cos \theta}

Given the equation: (3cos⁡θ−tan⁡θ)cos⁡θ=2 (3 \cos \theta - \tan \theta) \cos \theta = 2 we substitute tan⁡θ\tan \theta with sin⁡θcos⁡θ\frac{\sin \theta}{\cos \theta}: (3cos⁡θ−sin⁡θcos⁡θ)cos⁡θ=2 (3 \cos \theta - \frac{\sin \theta}{\cos \theta}) \cos \theta = 2

Step 2: Simplify the Equation

Simplify the equation by distributing cos⁡θ\cos \theta: 3cos⁡2θ−sin⁡θ=2 3 \cos^2 \theta - \sin \theta = 2

Step 3: Express in Terms of sin⁡θ\sin \theta

Using the Pythagorean identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta, substitute cos⁡2θ\cos^2 \theta: 3(1−sin⁡2θ)−sin⁡θ=2 3 (1 - \sin^2 \theta) - \sin \theta = 2 Simplify the equation: 3−3sin⁡2θ−sin⁡θ=2 3 - 3 \sin^2 \theta - \sin \theta = 2

Step 4: Rearrange to Quadratic Form

Rearrange the equation to the form asin⁡2θ+bsin⁡θ+c=0a \sin^2 \theta + b \sin \theta + c = 0: −3sin⁡2θ−sin⁡θ+1=0 -3 \sin^2 \theta - \sin \theta + 1 = 0 Multiply through by −1-1 to ensure a>0a > 0: 3sin⁡2θ+sin⁡θ−1=0 3 \sin^2 \theta + \sin \theta - 1 = 0

Final Answer

3sin⁡2θ+sin⁡θ−1=0 \boxed{3 \sin^2 \theta + \sin \theta - 1 = 0}

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