Question

Number of value of $x\in [0,4\pi ]$ and satisfying $\sqrt{2}\sec x+\tan x=1$ is?

Answer 1

$Giventheequation,\sqrt{2}secx+tanx=1⟶\left(1\right)\Rightarrow \frac{\sqrt{2}}{cosx}+\frac{sinx}{cosx}=1\Rightarrow \frac{\sqrt{2}+sinx}{cosx}=1\Rightarrow \sqrt{2}+sinx=cosx\Rightarrow sinx-cosx=\sqrt{2}squaringbothsidesweget,\Rightarrow 1-2sinxcosx=2\Rightarrow 2sinxcosx=-1\Rightarrow sin2x=-1\Rightarrow 2x={sin}^{-1}(-1)\Rightarrow 2x=\frac{3\pi }{2},\frac{7\pi }{2}\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}forxϵ[0,4\pi ]\because thevaluesofxliebetween[0,4\pi ]thereforetheequation\left(1\right)hastwovalues.$