Question

√3 sec 2x + 2 = 0

Explanation:

Step1: Simplify the equation

We know that \(\sec x=\frac{1}{\cos x}\). The given equation is \(\sqrt{3}\sec2x + 2=0\), which can be rewritten as \(\sqrt{3}\frac{1}{\cos2x}+2 = 0\). Then \(\frac{\sqrt{3}}{\cos2x}=- 2\), and \(\cos2x=-\frac{\sqrt{3}}{2}\).

Step2: Solve for \(2x\)

We know that if \(\cos\theta =-\frac{\sqrt{3}}{2}\), then \(\theta=\frac{5\pi}{6}+2n\pi\) or \(\theta=\frac{7\pi}{6}+2n\pi\), \(n\in\mathbf{Z}\). Let \(\theta = 2x\), so \(2x=\frac{5\pi}{6}+2n\pi\) or \(2x=\frac{7\pi}{6}+2n\pi\), \(n\in\mathbf{Z}\).

Step3: Solve for \(x\)

Divide both sides of the equations \(2x=\frac{5\pi}{6}+2n\pi\) and \(2x=\frac{7\pi}{6}+2n\pi\) by \(2\). We get \(x=\frac{5\pi}{12}+n\pi\) or \(x=\frac{7\pi}{12}+n\pi\), \(n\in\mathbf{Z}\).

Answer:

\(x=\frac{5\pi}{12}+n\pi,\frac{7\pi}{12}+n\pi,n\in\mathbf{Z}\)