find all solutions in the interval $0,2\\pi)$. \n$\\cos ^{2}\\theta - 2\\cos \\theta - 1 = 0$ \nselect the…

find all solutions in the interval $0,2\\pi)$. \n$\\cos ^{2}\\theta - 2\\cos \\theta - 1 = 0$ \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice. \na. $x =$ (type your answer in radians. round to four decimal places as needed. use a comma to separate answers as needed.) \nb. there is no solution.

find all solutions in the interval $0,2\\pi)$. \n$\\cos ^{2}\\theta - 2\\cos \\theta - 1 = 0$ \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice. \na. $x =$ (type your answer in radians. round to four decimal places as needed. use a comma to separate answers as needed.) \nb. there is no solution.

Answer

Explanation:

Step1: Let ( t = \cos\theta )

The equation ( \cos^{2}\theta-2\cos\theta - 1=0 ) becomes ( t^{2}-2t - 1=0 )

Step2: Solve the quadratic equation ( t^{2}-2t - 1=0 )

Using the quadratic formula ( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} ), where ( a = 1 ), ( b=-2 ), ( c=-1 ) [ \begin{align*} t&=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-1)}}{2\times1}\ &=\frac{2\pm\sqrt{4 + 4}}{2}\ &=\frac{2\pm\sqrt{8}}{2}\ &=\frac{2\pm2\sqrt{2}}{2}\ &=1\pm\sqrt{2} \end{align*} ]

Step3: Analyze the values of ( t )

Since ( - 1\leqslant\cos\theta\leqslant1 ) For ( t = 1+\sqrt{2}\approx1 + 1.4142=2.4142>1 ) (rejected) For ( t=1-\sqrt{2}\approx1-1.4142=-0.4142 )

Step4: Find ( \theta ) when ( \cos\theta=1 - \sqrt{2} )

( \theta=\cos^{-1}(1-\sqrt{2})) and ( \theta = 2\pi-\cos^{-1}(1-\sqrt{2}) ) ( \cos^{-1}(1 - \sqrt{2})\approx2.0344 ) radians ( 2\pi-\cos^{-1}(1-\sqrt{2})\approx2\pi - 2.0344\approx4.2488 ) radians

Answer:

A. ( x = 2.0344,4.2488 )