Question

The general solution of the equation \[ 2\sin^2 \theta - \cos^2 \theta = \sin \theta \]

Hint:

When solving trigonometric equations involving \(\sin^2 \theta\) and \(\cos^2 \theta\), use the Pythagorean identity to convert into a quadratic equation in \(\sin \theta\) or \(\cos \theta\), then apply the quadratic formula.

Answer 1

Given the equation: \[ 2\sin^2 \theta - \cos^2 \theta = \sin \theta \] Step 1: Use the Pythagorean identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation: \[ 2\sin^2 \theta - (1 - \sin^2 \theta) = \sin \theta \] \[ 2\sin^2 \theta - 1 + \sin^2 \theta = \sin \theta \] \[ 3 \sin^2 \theta - 1 = \sin \theta \] Step 2: Rearrange the equation to standard quadratic form in terms of \(\sin \theta\): \[ 3 \sin^2 \theta - \sin \theta - 1 = 0 \] Step 3: Let \(x = \sin \theta\), then solve: \[ 3x^2 - x - 1 = 0 \] Step 4: Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=3\), \(b=-1\), \(c=-1\): \[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 3 \times (-1)}}{2 \times 3} = \frac{1 \pm \sqrt{1 + 12}}{6} = \frac{1 \pm \sqrt{13}}{6} \] Step 5: Calculate the two roots: \[ x_1 = \frac{1 + \sqrt{13}}{6} \approx 0.7676 \] \[ x_2 = \frac{1 - \sqrt{13}}{6} \approx -0.4343 \] Both values lie within the valid range of \(\sin \theta\) (i.e., \(-1 \leq \sin \theta \leq 1\)).
Step 6: Find the general solution for each root: For \(\sin \theta = 0.7676\), \[ \theta = \sin^{-1}(0.7676) + 2n\pi \quad \text{or} \quad \pi - \sin^{-1}(0.7676) + 2n\pi, \] where \(n \in \mathbb{Z}\). Calculate \(\sin^{-1}(0.7676) \approx 0.876\) radians. So, \[ \theta = 0.876 + 2n\pi \quad \text{or} \quad 2.266 + 2n\pi. \] For \(\sin \theta = -0.4343\), \[ \theta = \sin^{-1}(-0.4343) + 2n\pi \quad \text{or} \quad \pi - \sin^{-1}(-0.4343) + 2n\pi. \] Calculate \(\sin^{-1}(-0.4343) \approx -0.448\) radians. So, \[ \theta = -0.448 + 2n\pi \quad \text{or} \quad 3.590 + 2n\pi. \] Thus, the general solutions are: \[ \boxed{ \begin{cases} \theta = 0.876 + 2n\pi, \quad \theta = 2.266 + 2n\pi, \\ \theta = -0.448 + 2n\pi, \quad \theta = 3.590 + 2n\pi, \end{cases} \quad n \in \mathbb{Z}. } \]