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

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). Let (t=\sin(x)), then the quadratic equation is (8t^{2}+6t + 1=0).

Step2: Solve the quadratic equation

For a quadratic equation (at^{2}+bt + c = 0) ((a = 8), (b = 6), (c = 1)), the quadratic formula is (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). 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}). Since (t = \sin(x)), when (\sin(x)=-\frac{1}{2}), (x=\frac{7\pi}{6}\approx3.67) or (x=\frac{11\pi}{6}\approx5.76) (because (\sin(x)) has a period of (2\pi) and (\sin(\frac{7\pi}{6})=\sin(\frac{11\pi}{6})=-\frac{1}{2})). 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(x)=-\alpha) gives (x=\pi+\arcsin(\alpha)) or (x = 2\pi-\arcsin(\alpha)) for (0\leq x<2\pi)).

Answer:

(3.49,3.67,5.76,5.94)