Question

solving trigonometric equations vi
solving trigonometric equations
part vi

solve ( 8 sin ^ { 2 } ( x ) - 6 cos ( x ) - 9 = 0 ) for all solutions ( 0 leq x < 2 pi )

( x = )

give your answers accurate to 2 decimal places, as a list separated by commas

question help: video

Explanation:

Step1: Use Pythagorean identity

Since \(\sin^{2}x = 1-\cos^{2}x\), the equation \(8\sin^{2}(x)-6\cos(x)-9 = 0\) becomes \(8(1 - \cos^{2}x)-6\cos x-9=0\).
Expand: \(8 - 8\cos^{2}x-6\cos x - 9=0\).
Simplify to: \(8\cos^{2}x+6\cos x + 1 = 0\).

Step2: Let \(t=\cos x\)

The quadratic equation is \(8t^{2}+6t + 1=0\).
Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 8\), \(b = 6\), \(c = 1\).
\(t=\frac{-6\pm\sqrt{36 - 32}}{16}=\frac{-6\pm2}{16}\).
So \(t_1=\frac{-6 + 2}{16}=-\frac{1}{4}\), \(t_2=\frac{-6-2}{16}=-\frac{1}{2}\).

Step3: Solve for \(x\)

When \(\cos x=-\frac{1}{2}\), \(x=\frac{2\pi}{3}\approx2.09\) or \(x=\frac{4\pi}{3}\approx4.19\).
When \(\cos x=-\frac{1}{4}\), \(x=\cos^{-1}(-\frac{1}{4})\approx1.82\) or \(x = 2\pi-\cos^{-1}(-\frac{1}{4})\approx4.46\).

Answer:

\(1.82,2.09,4.19,4.46\)