Solving a Trigonometric Equation In Exercises 25-38, find all solutions of the equation in the interval \( \mathbf{0 , 2 \pi )} \). \( \begin{array}{ll}\text { 25. } \cos ^{3} x=\cos x & \text { 26. } \sec ^{2} x-1=0\end{array} \)
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To solve \( \cos^{3} x = \cos x \), we can factor it as \( \cos x (\cos^{2} x - 1) = 0 \). This gives us two cases: \( \cos x = 0 \) which leads to \( x = \frac{\pi}{2}, \frac{3\pi}{2} \), and \( \cos^{2} x - 1 = 0 \) leads to \( \cos x = 1 \) or \( \cos x = -1 \), which gives us \( x = 0, \pi \). Combining, the solutions in the interval \( [0, 2\pi) \) are \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). For the equation \( \sec^{2} x - 1 = 0 \), we can rewrite it as \( \sec^{2} x = 1 \). Since \( \sec x = \frac{1}{\cos x} \), this implies \( \cos x = \pm 1 \). Therefore, the solutions are \( x = 0 \) (for \( \cos x = 1 \)) and \( x = \pi \) (for \( \cos x = -1 \)). Thus the solutions in the interval \( [0, 2\pi) \) are \( x = 0, \pi \).