5) (a) Show that the equation cos2x=sinx can be written in the form 2sin^2x+sinx-1=0 (b) Hence,solve cos2x=sinx where -pi leqslant xleqslant pi

(a) To show that the equation \( \cos 2x = \sin x \) can be written in the form \( 2\sin^2 x + \sin x - 1 = 0 \), we start by using the double-angle identity for cosine:<br /><br />\[ \cos 2x = 2\cos^2 x - 1 \]<br /><br />Given \( \cos 2x = \sin x \), we substitute \( \cos 2x \) with \( \sin x \):<br /><br />\[ \sin x = 2\cos^2 x - 1 \]<br /><br />Next, we use the Pythagorean identity \( \cos^2 x = 1 - \sin^2 x \) to express \( \cos^2 x \) in terms of \( \sin x \):<br /><br />\[ \cos^2 x = 1 - \sin^2 x \]<br /><br />Substitute \( \cos^2 x \) into the equation:<br /><br />\[ \sin x = 2(1 - \sin^2 x) - 1 \]<br /><br />Simplify the equation:<br /><br />\[ \sin x = 2 - 2\sin^2 x - 1 \]<br /><br />\[ \sin x = 1 - 2\sin^2 x \]<br /><br />Rearrange the equation to the standard quadratic form:<br /><br />\[ 2\sin^2 x + \sin x - 1 = 0 \]<br /><br />Thus, the equation \( \cos 2x = \sin x \) can be written in the form \( 2\sin^2 x + \sin x - 1 = 0 \).<br /><br />(b) To solve \( \cos 2x = \sin x \) where \( -\pi \leq x \leq \pi \), we solve the quadratic equation \( 2\sin^2 x + \sin x - 1 = 0 \).<br /><br />Let \( u = \sin x \). The quadratic equation becomes:<br /><br />\[ 2u^2 + u - 1 = 0 \]<br /><br />We solve this quadratic equation using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -1 \):<br /><br />\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]<br /><br />\[ u = \frac{-1 \pm \sqrt{1 + 8}}{4} \]<br /><br />\[ u = \frac{-1 \pm \sqrt{9}}{4} \]<br /><br />\[ u = \frac{-1 \pm 3}{4} \]<br /><br />So, the solutions for \( u \) are:<br /><br />\[ u = \frac{1}{2} \quad \text{and} \quad u = -1 \]<br /><br />Now, we find the corresponding values of \( x \):<br /><br />1. For \( \sin x = \frac{1}{2} \):<br /><br />\[ x = \frac{\pi}{6} \quad \text{or} \quad x = \frac{5\pi}{6} \]<br /><br />2. For \( \sin x = -1 \):<br /><br />\[ x = -\frac{\pi}{2} \]<br /><br />Thus, the solutions to \( \cos 2x = \sin x \) in the interval \( -\pi \leq x \leq \pi \) are:<br /><br />\[ x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}, \quad x = -\frac{\pi}{2} \]