Answer to Question #194265 in Trigonometry for Bill

Question

A water wheel rotates through the angle x, the water level L behind the wheel changes according the equation "L = 1 - cos x - 2 sin^2 x" where L is measured in inches. Determine all values of x in degrees for which the water level is zero.

Answer 1

$L=1−cosx−2sin^2x$

$Put L=0 \space in \space L=1−cosx−2sin^2x, we \space get$

$⇒1−cosx−2sin^2x=0$

$⇒1−cosx−2(1−cos^2x)=0 \space \space [∵sin^2x+cos^2x=1]\\⇒1−cosx−2+2cos2x=0$

$⇒2cos^2x−cosx−1=0\\⇒2cos^2x−2cosx+cosx−1=0$

$⇒2cosx(cosx−1)+(cosx−1)=0\\⇒(cosx−1)(2cosx+1)=0\\⇒cosx−1=0 or 2cosx+1=0$

$⇒cosx=1\space or\space 2cosx=−1\\⇒cosx=1\space or \space cosx=−1/2\\⇒x=360^{\circ}n \,or \\ \,x=120^{\circ}+360^{\circ}n \space or\space x=240^{\circ}+360^{\circ}n, n∈ℤ$

Hence the values of x for which the water level is zero are,

$x=360^{\circ}n\space or\space x=120^{\circ}+360^{\circ}n\space or \space x=240^{\circ}+360^{\circ}n, n∈ℤ$