Question

General solution of the equation $4\cos x-3\sec x=\tan x,(\cos x\ne 0)$ can be

• A. $n\pi +(-1{)}^n\alpha ,\sin \alpha =\frac{-1+\sqrt{17}}{8},n\in Z$
• B. $n\pi +(-1{)}^n\beta ,\sin \beta =\frac{-1-\sqrt{17}}{8},n\in Z$
• C. $n\pi +(-1{)}^n\alpha ,\sin \alpha =\frac{1+\sqrt{17}}{8},n\in Z$
• D. $n\pi +(-1{)}^n\beta ,\sin \beta =\frac{1-\sqrt{17}}{8},n\in Z$

Answer 1

Answer: (A), (B)

The correct options are
A $n\pi +(-1{)}^n\alpha ,\sin \alpha =\frac{-1+\sqrt{17}}{8},n\in Z$
B $n\pi +(-1{)}^n\beta ,\sin \beta =\frac{-1-\sqrt{17}}{8},n\in Z$

$4\cos x-3\sec x=\tan x\Rightarrow 4\cos x-\frac{3}{\cos x}=\frac{\sin x}{\cos x}\Rightarrow 4{\cos }^{2}x-3=\sin x\Rightarrow 4(1-{\sin }^{2}x)-3=\sin x\Rightarrow 4{\sin }^{2}x+\sin x-1=0\Rightarrow \sin x=\frac{-1\pm \sqrt{17}}{8}$
If $\sin x=\frac{-1+\sqrt{17}}{8}$
$\therefore x=n\pi +(-1{)}^n\alpha ,\sin \alpha =\frac{-1+\sqrt{17}}{8},n\in Z$
If $\sin x=\frac{-1-\sqrt{17}}{8}$
$\therefore x=n\pi +(-1{)}^n\beta ,\sin \beta =\frac{-1-\sqrt{17}}{8},n\in Z$