Question

The solutions of the equation \(cos\theta=2-3sin(\frac{\theta}{2})\ in\ the\ interval\ 0\leq\theta\leq\pi\ are\)

• A. \(\frac{\pi}{4},\pi\)
• B. \(\frac{\pi}{3},\frac{\pi}{2}\)
• C. \(\frac{\pi}{3},\pi\)
• D. \(\frac{\pi}{6},\frac{\pi}{2}\)
• E. \(\frac{\pi}{6},\pi\)

Answer 1

The Correct Option is C

We are given the equation $\cos \theta = 2 - 3\sin \left( \frac{\theta}{2} \right)$ in the interval $0 \leq \theta \leq \pi$.
To solve it, we first need to express $\cos \theta$ in terms of $\sin \left( \frac{\theta}{2} \right)$.
Let $\alpha = \frac{\theta}{2}$, so that $\theta = 2\alpha$.
The equation becomes: \[ \cos(2\alpha) = 2 - 3\sin(\alpha) \] Using the identity for $\cos(2\alpha)$: \[ \cos(2\alpha) = 1 - 2\sin^2(\alpha) \] The equation becomes: \[ 1 - 2\sin^2(\alpha) = 2 - 3\sin(\alpha) \] Rearranging the equation: \[ 2\sin^2(\alpha) - 3\sin(\alpha) - 1 = 0 \] This is a quadratic equation in $\sin(\alpha)$. Using the quadratic formula: \[ \sin(\alpha) = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)} \] \[ \sin(\alpha) = \frac{3 \pm \sqrt{9 + 8}}{4} = \frac{3 \pm \sqrt{17}}{4} \] The possible values of $\sin(\alpha)$ are approximately: \[ \sin(\alpha) = 1 \quad \text{or} \quad \sin(\alpha) \approx -0.5 \] For the first value, $\alpha = \frac{\pi}{2}$, so $\theta = 2\alpha = \pi$. For the second value, $\alpha = \frac{\pi}{6}$, so $\theta = 2\alpha = \frac{\pi}{3}$.

The correct option is (C) : \(\frac{\pi}{3},\pi\)

Answer 2

We are given the equation \(\cos\theta = 2 - 3\sin\left(\frac{\theta}{2}\right)\) and we want to find the solutions in the interval \(0 \le \theta \le \pi\).

We can use the identity \(\cos\theta = 1 - 2\sin^2\left(\frac{\theta}{2}\right)\). Substituting this into the equation, we get:

\(1 - 2\sin^2\left(\frac{\theta}{2}\right) = 2 - 3\sin\left(\frac{\theta}{2}\right)\)

Let \(x = \sin\left(\frac{\theta}{2}\right)\). The equation becomes:

\(1 - 2x^2 = 2 - 3x\)

Rearranging the terms, we get a quadratic equation:

\(2x^2 - 3x + 1 = 0\)

We can factor this quadratic equation:

\((2x - 1)(x - 1) = 0\)

This gives us two possible solutions for \(x\):

\(x = \frac{1}{2}\) or \(x = 1\)

Since \(x = \sin\left(\frac{\theta}{2}\right)\), we have:

\(\sin\left(\frac{\theta}{2}\right) = \frac{1}{2}\) or \(\sin\left(\frac{\theta}{2}\right) = 1\)

For \(\sin\left(\frac{\theta}{2}\right) = \frac{1}{2}\), we have \(\frac{\theta}{2} = \frac{\pi}{6}\) (since \(0 \le \frac{\theta}{2} \le \frac{\pi}{2}\)). This gives us \(\theta = \frac{\pi}{3}\).

For \(\sin\left(\frac{\theta}{2}\right) = 1\), we have \(\frac{\theta}{2} = \frac{\pi}{2}\). This gives us \(\theta = \pi\).

Both solutions are in the interval \(0 \le \theta \le \pi\). Therefore, the solutions are \(\theta = \frac{\pi}{3}\) and \(\theta = \pi\).