Question

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

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

Answer 1

Answer: (A), (D)

$x=n\pi +(-1{)}^n\beta ,\beta ={\sin }^{-1}\left(\frac{-1-\sqrt{17}}{8}\right),n\in Z$

$4\cos x-3\sec x=\tan x\Rightarrow 4\cos x-\frac{3}{\cos x}=\frac{\sin x}{\cos x}\Rightarrow 4{\cos }^{2}x-3=\sin x\Rightarrow 4(1-{\sin }^{2}x)-3=\sin x\Rightarrow 4{\sin }^{2}x+\sin x-1=0\Rightarrow \sin x=\frac{-1\pm \sqrt{17}}{8}$

If $\sin x=\frac{-1+\sqrt{17}}{8},$ then
$x=n\pi +(-1{)}^n\alpha ,\alpha ={\sin }^{-1}\left(\frac{-1+\sqrt{17}}{8}\right),n\in Z$

If $\sin x=\frac{-1-\sqrt{17}}{8},$ then
$x=n\pi +(-1{)}^n\beta ,\beta ={\sin }^{-1}\left(\frac{-1-\sqrt{17}}{8}\right),n\in Z$