Question

Q21 : Do the question of finding the general solution of follwing trigonometric equations

Q21. Find the general solution of following trigonometric equation:
${\text{cos}}^{2}xcosecx+3sinx+3=0$

Answer 1

Dear student
$$\begin{gathered} {\cos }^2\mathrm{x}\mathrm{cosecx}+3\mathrm{sinx}+3=0 \\ \Rightarrow \left(1-{\sin }^2\mathrm{x}\right)\mathrm{cosecx}+3\left(\mathrm{sinx}+1\right)=0 \\ \Rightarrow \left(\mathrm{sinx}+1\right)\left[-\left(\mathrm{sinx}-1\right)\mathrm{cosecx}+3\right]=0 \\ \Rightarrow \mathrm{sinx}+1=0\mathrm{or}\left[-\left(\mathrm{sinx}-1\right)\mathrm{cosecx}+3\right]=0 \\ \mathrm{Consider}, \\ \mathrm{sinx}+1=0 \\ \Rightarrow \mathrm{sinx}=-1 \\ \Rightarrow \mathrm{x}=\frac{3\mathrm{\pi }}{2}+2\mathrm{n\pi },\mathrm{n}\in \mathrm{Z} \\ \mathrm{and}\left[-\left(\mathrm{sinx}-1\right)\mathrm{cosecx}+3\right]=0 \\ \Rightarrow \frac{2\mathrm{sinx}+1}{\mathrm{sinx}}=0 \\ \Rightarrow 1+2\mathrm{sinx}=0 \\ \Rightarrow \mathrm{sinx}=\frac{-1}{2} \\ \Rightarrow \mathrm{x}=\frac{7\mathrm{\pi }}{6}+2\mathrm{n\pi },\mathrm{x}=\frac{11\mathrm{\pi }}{6}+2\mathrm{n\pi } \\ \mathrm{So},\mathrm{x}=\frac{3\mathrm{\pi }}{2}+2\mathrm{n\pi },\mathrm{n}\in \mathrm{Z},\mathrm{x}=\frac{7\mathrm{\pi }}{6}+2\mathrm{n\pi },\mathrm{x}=\frac{11\mathrm{\pi }}{6}+2\mathrm{n\pi } \end{gathered}$$
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