Question

assignment 7.5: solving trigonometric equations
score: 30/100 answered: 3/10
question 4
solve \\( \sin ^ { 2 } ( x ) = - 6 \cos ( x ) \\) for all solutions \\( 0 \leq x < 2 \pi \\).
\\( x = \\)
give your answers as values accurate to at least two decimal places in a list separated by commas.
question help: video 1 video 2 post to forum

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:

\(1.73,4.55\)