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.

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 \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).
2. Substitute \(\tan \theta\) and simplify the equation.
3. Rearrange the equation to match the form \(a \sin^2 \theta + b \sin \theta + c = 0\).

Given the equation: \[ (3 \cos \theta - \tan \theta) \cos \theta = 2 \] we substitute \(\tan \theta\) with \(\frac{\sin \theta}{\cos \theta}\): \[ (3 \cos \theta - \frac{\sin \theta}{\cos \theta}) \cos \theta = 2 \]

Simplify the equation by distributing \(\cos \theta\): \[ 3 \cos^2 \theta - \sin \theta = 2 \]

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

Rearrange the equation to the form \(a \sin^2 \theta + b \sin \theta + c = 0\): \[ -3 \sin^2 \theta - \sin \theta + 1 = 0 \] Multiply through by \(-1\) to ensure \(a > 0\): \[ 3 \sin^2 \theta + \sin \theta - 1 = 0 \]

Final Answer