Question

The general solution of the trigonometric equation $\sin x+\cos x=1$ is given by:

• A. $x=2n\pi ;n=0,\pm 1,\pm 2...$
• B. $x=2n\pi +\pi ;n=0,\pm 1,\pm 2...$
• C. $x=n\pi +(-1{)}^n\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4};n=0,\pm 1,\pm 2...$
• D. none of these

Answer 1

The correct option is C

$x=n\pi +(-1{)}^n\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4};n=0,\pm 1,\pm 2...$

Step 1: Express as Trigonometric Identity

Given,

$\sin x+\cos x=1$

Multiplying and dividing LHS by $\sqrt{2}$,

$\sqrt{2}\left(\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}\right)=1$

$\Rightarrow$$\sin x\sin \frac{\mathrm{\pi }}{4}+\cos x\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}$

Using the formula $\sin (A+B)=\sin A\cos B+\cos A\sin B$,

$\Rightarrow$$\sin \left(x+\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}$

Step 2: Estimate the General Equation

Therefore,

$\sin \left(x+\frac{\mathrm{\pi }}{4}\right)=\sin \left[n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4}\right]$

where $n=0,\pm 1,\pm 2...$

$\Rightarrow$$x+\frac{\mathrm{\pi }}{4}=n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4}$

$\Rightarrow$ $x=n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}$

Hence, the correct answer is option (C).