chapter 7 quiz\n150 points possible answered: 10/15\nquestion 11\nsolve ( 8 cos ^{2}(x)-6 sin (x)-9=0 ) for…

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)