Question

Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].

Answer 1

Given:
tanx + secx = 2 cosx

$$\begin{gathered} \Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x \\ \Rightarrow \frac{\sin x+1}{\cos x}=2\cos x \\ \Rightarrow \sin x+1=2{\cos }^{2}x \\ \Rightarrow \sin x=2{\cos }^{2}x-1 \end{gathered}$$

$$\begin{gathered} \Rightarrow 2\left(1-{\sin }^{2}x\right)-1=\sin x \\ \Rightarrow 2-2{\sin }^{2}x-1=\sin x \\ \Rightarrow 1-2{\sin }^{2}x=\sin x \\ \Rightarrow 2{\sin }^{2}x+\sin x-1=0 \\ \Rightarrow 2{\sin }^{2}x+2\sin x-\sin x-1=0 \\ \Rightarrow 2\sin x\left(\sin x+1\right)-1\left(\sin x+1\right)=0 \\ \Rightarrow \left(\sin x+1\right)\left(2\sin x-1\right)=0 \\ \Rightarrow \sin x+1=0\mathrm{or}2\sin x-1=0 \\ \Rightarrow \sin x=-1\mathrm{or}\sin x=\frac{1}{2} \end{gathered}$$
Now,
$$\begin{gathered} \sin x=-1 \\ \Rightarrow \sin x=\sin \left(\frac{3\mathrm{\pi }}{2}\right) \\ \Rightarrow x=n\mathrm{\pi }+{\left(-1\right)}^n\frac{3\mathrm{\pi }}{2},n\in Z \\ \mathrm{Because}\mathrm{it}\mathrm{contains}\mathrm{an}\mathrm{odd}\mathrm{multiple}\mathrm{of}\frac{\mathrm{\pi }}{2}\mathrm{and}\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{tanx}\mathrm{and}\mathrm{secx}\mathrm{are}\mathrm{undefined}\mathrm{on}\mathrm{the}\mathrm{odd}\mathrm{multiple},\mathrm{this}\mathrm{value}\mathrm{will}\mathrm{not}\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{equation}. \end{gathered}$$
And,
$$\begin{gathered} \sin x=\frac{1}{2} \\ \Rightarrow \sin x=\sin \left(\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow x=n\mathrm{\pi }+{\left(-1\right)}^n\frac{\mathrm{\pi }}{6},n\in Z \\ \mathrm{Now}, \\ \mathrm{For}n=0,x=\frac{\mathrm{\pi }}{6} \\ \mathrm{For}n=1,x=\frac{11\mathrm{\pi }}{6} \\ \mathrm{For}\mathrm{other}\mathrm{values}\mathrm{of}n,\mathrm{the}\mathrm{condition}\mathrm{is}\mathrm{not}\mathrm{true}. \end{gathered}$$

Hence, the given equation has two solutions in $\left[0,2\mathrm{\pi }\right]$.