Chapter 5.4, Problem 7QR

Solution

To find: The general solution of the given equation.

The general solutions of $\left(2\sin x-1\right)\left(2\cos x+1\right)=0$ are $\frac{\pi }{6}\text{+2}\pi \text{n ,}\frac{5\pi }{6}\text{ +2}\pi \text{n}$

  $\frac{2\pi }{3}\text{+2}\pi \text{n and }\frac{4\pi }{3}+2\pi n$ .

Given data:

Equation: $\left(2\sin x-1\right)\left(2\cos x+1\right)=0$

Formula used:

Sine and cosine function are a periodic functions with period $2\pi$ .

Calculation:

  $\left(2\sin x-1\right)\left(2\cos x+1\right)=0$

Further solve the equation.

  $\begin{array}{l}2\sin x-1=0or2\cos x+1=0\\ \sin x=\frac{1}{2}or\cos x=\frac{-1}{2}\\ x={\sin }^{-1}\left(\frac{1}{2}\right)orx={\cos }^{-1}\left(\frac{-1}{2}\right)\\ x=\frac{\pi }{6},\frac{5\pi }{6}orx=\frac{2\pi }{3},\frac{4\pi }{3}\end{array}$

The solutions of $2\sin x-1=0$ are $\frac{\pi }{6}\text{ and }\frac{5\pi }{6}$ . The general solutions can be obtained by periodicity that are $\frac{\pi }{6}\text{+2}\pi \text{n and }\frac{5\pi }{6}\text{ +2}\pi \text{n}$ .

The solutions of $2\cos x+1=0$ are $\frac{2\pi }{3}\text{ and }\frac{4\pi }{3}$ . The general solutions can be obtained by periodicity that are $\frac{2\pi }{3}\text{+2}\pi \text{n and }\frac{4\pi }{3}+2\pi n$ .

Therefore, the general solutions of $\left(2\sin x-1\right)\left(2\cos x+1\right)=0$ are $\frac{\pi }{6}\text{+2}\pi \text{n ,}\frac{5\pi }{6}\text{ +2}\pi \text{n}$ $\frac{2\pi }{3}\text{+2}\pi \text{n and }\frac{4\pi }{3}+2\pi n$ .