question 8 solve sin²(x) = -8 cos(x) for all solutions 0 ≤ x < 2π x = give your answers accurate to 2…

Explanation:

Step1: Use trig identity

Recall $\sin^{2}(x)=1 - \cos^{2}(x)$. So the equation becomes $1-\cos^{2}(x)=- 8\cos(x)$.

Step2: Rearrange to quadratic form

Let $u = \cos(x)$, then $u^{2}-8u - 1=0$.

Step3: Apply quadratic formula

The quadratic formula for $au^{2}+bu + c = 0$ is $u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here $a = 1$, $b=-8$, $c=-1$. So $u=\frac{8\pm\sqrt{(-8)^{2}-4\times1\times(-1)}}{2\times1}=\frac{8\pm\sqrt{64 + 4}}{2}=\frac{8\pm\sqrt{68}}{2}=\frac{8\pm2\sqrt{17}}{2}=4\pm\sqrt{17}$.

Step4: Determine valid values of u

Since $-1\leqslant\cos(x)\leqslant1$, $u = 4+\sqrt{17}\approx4 + 4.123=8.123$ is not valid. So $u=\cos(x)=4-\sqrt{17}\approx4 - 4.123=-0.123$.

Step5: Find x values

$x=\cos^{-1}(-0.123)$. In the range $0\leqslant x<2\pi$, $x\approx1.79,4.49$.

Answer:

$1.79,4.49$