Question

(1) Solve sin x cos x = 0.

(2) Solve 2 sin2 x − sin x − 1 = 0.

(3) Solve cos x + cos 2x = 0.

(4) Solve 2 cos x + sec x = 3.

(5) Solve sec x − 1 = tan x.

Answer 1

Answer on Question #75946 – Math – Trigonometry

Question

(1) $\sin x \cos x = 0$

Solution

$$\begin{array}{l} \sin x \cos x = 0 \\ \frac{1}{2} 2 \sin x \cos x = 0 \\ \frac{1}{2} \sin 2x = 0 \\ \sin 2x = 0 \\ 2x = \pi k, k \in \mathbb{Z} \\ x = \frac{1}{2} \pi k, k \in \mathbb{Z} \\ \end{array}$$

Answer: $x = \frac{1}{2} \pi k, k \in \mathbb{Z}$

Question

(2) $2 \sin^2 x - \sin x - 1 = 0$

Solution

$$\begin{array}{l} 2 \sin^2 x - \sin x - 1 = 0 \\ 2 \sin^2 x - 2 \sin x + \sin x - 1 = 0 \\ 2 \sin x (\sin x - 1) + (\sin x - 1) = 0 \\ (2 \sin x + 1) (\sin x - 1) = 0 \\ a) 2 \sin x + 1 = 0 \quad \vee \quad b) \sin x - 1 = 0 \\ \end{array}$$

a) $2 \sin x + 1 = 0$

$$\begin{array}{l} 2 \sin x + 1 = 0 \\ \sin x = -\frac{1}{2} \\ x = (-1)^{k+1} \frac{\pi}{6} + \pi k, k \in \mathbb{Z} \\ \end{array}$$

b) $\sin x - 1 = 0$

$$\begin{array}{l} \sin x - 1 = 0 \\ \sin x = 1 \\ x = \frac{\pi}{2} + 2\pi k, k \in \mathbb{Z} \\ \end{array}$$

Answer: $x = (-1)^{k+1} \frac{\pi}{6} + \pi k, k \in \mathbb{Z}$ and $x = \frac{\pi}{2} + 2\pi k, k \in \mathbb{Z}$

Question

(3) $\cos x + \cos (2x) = 0$

Solution

$$\begin{array}{l} \cos x + \cos (2x) = 0 \\ \cos x + 2 \cos^2 x - 1 = 0 \\ (\cos x + 1)(2 \cos x - 1) = 0 \\ \end{array}$$

a) $\cos x + 1 = 0 \quad \vee \quad b) 2 \cos x - 1 = 0$

$$\begin{array}{l} \text{a) } \cos x + 1 = 0 \\ \cos x = -1 \\ x = \pi + 2\pi k, k \in \mathbb{Z} \end{array}$$

b) $2 \cos x - 1 = 0$

$$\begin{array}{l} \cos x = \frac{1}{2} \\ x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z} \end{array}$$

Answer: $x = \pi + 2\pi k, k \in \mathbb{Z}$ and $x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}$

Question

(4) $2 \cos x + \sec x = 3$

Solution

$$\begin{array}{l} 2 \cos x + \sec x = 3 \\ 2 \cos x + \frac{1}{\cos x} = 3 \\ 2 \cos^2 x + 1 = 3 \cos x \\ 2 \cos^2 x - 3 \cos x + 1 = 0 \\ \text{Let } t = \cos x \\ 2t^2 - 3t + 1 = 0 \\ D = 9 - 8 = 1 \\ t_1 = \frac{3 - 1}{4} = \frac{1}{2} \\ t_2 = \frac{3 + 1}{4} = 1 \\ \end{array}$$

So a) $\cos x = \frac{1}{2}$ or b) $\cos x = 1$

$$\begin{array}{l} \text{a) } \cos x = \frac{1}{2} \\ x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z} \end{array}$$

b) $\cos x = 1$

$$x = 2\pi k, k \in \mathbb{Z}$$

Answer: $x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}; x = 2\pi k, k \in \mathbb{Z}$

Question

(5) $\sec x - 1 = \tan x$

Solution

$$\sec x - 1 = \tan x$$ $$\begin{array}{l} \frac{1}{\cos x} - 1 = \tan x \\ 1 - \cos x = \frac{\sin x}{\cos x} \cos x \\ \sin x + \cos x = 1 \\ \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}} \\ \sin \left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \\ x = - \frac{\pi}{4} + (-1)^k \frac{\pi}{4} + \pi k, \, k \in \mathbb{Z} \end{array}$$

Answer: $x = -\frac{\pi}{4} + (-1)^k \frac{\pi}{4} + \pi k, \, k \in \mathbb{Z}$.

Answer provided by https://www.AssignmentExpert.com