solve 8 sin²(x) - 2 cos(x) - 5 = 0 for all solutions 0 ≤ x < 2π\nx =\ngive your answers accurate to 2…

Explanation:

Step1: Use the identity $\sin^{2}(x)=1 - \cos^{2}(x)$

Substitute $\sin^{2}(x)$ in the given equation: $8(1 - \cos^{2}(x))-2\cos(x)-5 = 0$ Expand to get $8-8\cos^{2}(x)-2\cos(x)-5 = 0$ Rearrange to $8\cos^{2}(x)+2\cos(x)-3 = 0$

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

The equation becomes a quadratic equation $8t^{2}+2t - 3=0$ Use the quadratic - formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $at^{2}+bt + c = 0$. Here $a = 8$, $b = 2$, $c=-3$ $t=\frac{-2\pm\sqrt{2^{2}-4\times8\times(-3)}}{2\times8}=\frac{-2\pm\sqrt{4 + 96}}{16}=\frac{-2\pm\sqrt{100}}{16}=\frac{-2\pm10}{16}$

Step3: Solve for $t$

$t_1=\frac{-2 + 10}{16}=\frac{8}{16}=\frac{1}{2}$ $t_2=\frac{-2-10}{16}=\frac{-12}{16}=-\frac{3}{4}$

Step4: Substitute back $t = \cos(x)$

When $\cos(x)=\frac{1}{2}$, $x=\frac{\pi}{3}, \frac{5\pi}{3}$ (since $0\leq x\lt2\pi$) When $\cos(x)=-\frac{3}{4}$, $x=\cos^{-1}(-\frac{3}{4})\approx2.42, 3.86$ (using a calculator to find the inverse - cosine values in the range $0\leq x\lt2\pi$)

Answer:

$1.05,2.42,3.86,5.24$