assignment 7.5: solving trigonometric equations\nscore: 30/100 answered: 3/10\nquestion 4\nsolve \\( \\sin ^…

Explanation:

Step1: Use Pythagorean identity

We know that (\sin^{2}(x)=1 - \cos^{2}(x)). Substitute this into the equation (\sin^{2}(x)=-6\cos(x)), we get (1-\cos^{2}(x)=-6\cos(x)).

Step2: Rearrange the equation

Rearrange the equation (1-\cos^{2}(x)=-6\cos(x)) to the quadratic form. Let (t = \cos(x)), then the equation becomes (t^{2}-6t - 1=0).

Step3: Solve the quadratic equation

Use the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (at^{2}+bt + c = 0). Here (a = 1), (b=-6), (c=-1). So (t=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times(-1)}}{2\times1}=\frac{6\pm\sqrt{36 + 4}}{2}=\frac{6\pm\sqrt{40}}{2}=\frac{6\pm2\sqrt{10}}{2}=3\pm\sqrt{10}).

Step4: Find the valid value of (\cos(x))

Since (- 1\leqslant\cos(x)\leqslant1), (t = 3+\sqrt{10}\approx3 + 3.16=6.16) (rejected). (t=3-\sqrt{10}\approx3-3.16=-0.16). So (\cos(x)=3 - \sqrt{10}\approx - 0.16).

Step5: Find (x) values

If (\cos(x)=a), then (x=\cos^{-1}(a)+2k\pi) or (x = 2\pi-\cos^{-1}(a)+2k\pi), (k\in\mathbb{Z}). For (0\leqslant x<2\pi), (x=\cos^{-1}(-0.16)\approx1.73) and (x = 2\pi-\cos^{-1}(-0.16)\approx4.55).

Answer:

(1.73,4.55)