Complete solutions to trigonometric equation problems are presented below. In all solutions, \( n \) represents any integer and \( \pi \) represents the radian measure of a straight angle (\(180^\circ\)).
Solve for all solutions:
\[ 2 \cos x + 1 = 0 \]
Answer: \( \displaystyle x = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2n\pi \)
Solve for all solutions:
\[ 3 \sec^2 x - 4 = 0 \]
Answer: \( \displaystyle x = \frac{\pi}{6} + 2n\pi, \quad x = \frac{5\pi}{6} + 2n\pi, \quad x = \frac{7\pi}{6} + 2n\pi, \quad x = \frac{11\pi}{6} + 2n\pi \)
Note: This includes all four families of solutions from \( \sec x = \pm \frac{2}{\sqrt{3}} \).
Solve for all solutions:
\[ (3 \cos x + 7)(-2 \sin x - 1) = 0 \]
Answer: \( \displaystyle x = \frac{7\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{11\pi}{6} + 2n\pi \)
Note: The equation \( 3\cos x + 7 = 0 \) has no solution since \( \cos x = -\frac{7}{3} \) is outside the range \([-1, 1]\).
Solve for all solutions:
\[ (6 \tan^2 x - 2)(2 \tan^2 x - 6) = 0 \]
Answer: \( \displaystyle x = \frac{\pi}{6} + n\pi, \quad x = \frac{5\pi}{6} + n\pi, \quad x = \frac{\pi}{3} + n\pi, \quad x = \frac{2\pi}{3} + n\pi \)
Solve in the interval \([0, 2\pi)\):
\[ -2 \sec^2 x + 4 = -2\sec x \]
Answer: \( \displaystyle x = \frac{\pi}{3}, \quad x = \pi, \quad x = \frac{5\pi}{3} \)
Solve in the interval \([0, 2\pi)\):
\[ 2\sin(x) \cos(-x) = 2 \sin(-x) \sin(x) \]
Answer: \( \displaystyle x = 0, \quad x = \frac{3\pi}{4}, \quad x = \pi, \quad x = \frac{7\pi}{4} \)
Solve in the interval \([0, 2\pi)\):
\[ \sin 2x = -\sin(-x) \]
Answer: \( \displaystyle x = 0, \quad x = \frac{\pi}{3}, \quad x = \pi, \quad x = \frac{5\pi}{3} \)
Which equation has no solution?
Answer: \( 2\sin x = -3 \)
Explanation: The range of sine is \([-1, 1]\), so \(2\sin x\) cannot equal \(-3\).
Which equation has no solution?
Answer: \( \sec x = -\frac{1}{2} \)
Explanation: The range of secant is \((-\infty, -1] \cup [1, \infty)\), so it cannot equal \(-\frac{1}{2}\).
Solve in the interval \([0, 2\pi)\):
\[ \sin^2 x + \sin x = 6 \]
Answer: No solutions
Explanation: The equation simplifies to \(\sin^2 x + \sin x - 6 = 0\), whose potential solutions \(\sin x = 2\) or \(\sin x = -3\) are outside the range of sine \([-1, 1]\).