Question

chapter 7 quiz
150 points possible answered: 10/15
question 11
solve ( 8 cos ^{2}(x)-6 sin (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
add work

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Explanation:

Step1: Use the Pythagorean identity

We know that \(\cos^{2}(x)=1 - \sin^{2}(x)\). Substitute this into the equation \(8\cos^{2}(x)-6\sin(x)-9 = 0\).
We get \(8(1-\sin^{2}(x))-6\sin(x)-9 = 0\).
Expand the left - hand side: \(8 - 8\sin^{2}(x)-6\sin(x)-9 = 0\).
Simplify to \( - 8\sin^{2}(x)-6\sin(x)-1 = 0\). Multiply through by \(-1\) to obtain \(8\sin^{2}(x)+6\sin(x)+1 = 0\).

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

The equation becomes a quadratic equation \(8t^{2}+6t + 1=0\).
Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 8\), \(b = 6\), and \(c = 1\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(6)^{2}-4\times8\times1=36 - 32 = 4\).
Then \(t=\frac{-6\pm\sqrt{4}}{2\times8}=\frac{-6\pm2}{16}\).
We have two solutions for \(t\):
\(t_{1}=\frac{-6 + 2}{16}=\frac{-4}{16}=-\frac{1}{4}\) and \(t_{2}=\frac{-6-2}{16}=\frac{-8}{16}=-\frac{1}{2}\).

Step3: Solve for \(x\) when \(\sin(x)=t\)

When \(\sin(x)=-\frac{1}{2}\), \(x=\frac{7\pi}{6}\approx3.67\) or \(x=\frac{11\pi}{6}\approx5.76\) (since \(\sin(x)\) has a period of \(2\pi\) and \(\sin(\frac{7\pi}{6})=\sin(\frac{11\pi}{6})=-\frac{1}{2}\) in the interval \(0\leq x\lt2\pi\)).
When \(\sin(x)=-\frac{1}{4}\), \(x=\pi+\arcsin(\frac{1}{4})\approx3.49\) or \(x = 2\pi-\arcsin(\frac{1}{4})\approx5.94\) (using the identity \(\sin(\pi+\alpha)=-\sin(\alpha)\) and \(\sin(2\pi-\alpha)=-\sin(\alpha)\)).

Answer:

\(3.49,3.67,5.76,5.94\)