NIMCET 2013 — Mathematics PYQ

To find the values of $x$ in the interval $[0,2\pi )$ that satisfy the equation, we follow these steps:

Step 1: Rearrange the equation

Step 2: Square both sides

To eliminate the trigonometric functions, square both sides of the equation:

Step 3: Simplify using identities

We know that ${\sin }^{2}x+{\cos }^{2}x=1$ and $2\sin x\cos x=\sin (2x)$. Substituting these:

Step 4: Find the general values for $2x$

For $\sin (2x)=0$ in the range corresponding to 0 \le x < 2\pi (which is 0 \le 2x < 4\pi):

Step 5: Verify the solutions

Since we squared the equation, we might have introduced "extraneous" solutions. We must check them in the original equation $\sin x+1=\cos x$:

• For $x=0$: $\sin (0)+1=0+1=1$; $\cos (0)=1$. (Matches: $1=1$) $✓$

• For $x=\frac{\pi }{2}$: $\sin (\frac{\pi }{2})+1=1+1=2$; $\cos (\frac{\pi }{2})=0$. (Does not match: $2\ne 0$) $\times$

• For $x=\pi$: $\sin (\pi )+1=0+1=1$; $\cos (\pi )=-1$. (Does not match: $1\ne -1$) $\times$

• For $x=\frac{3\pi }{2}$: $\sin (\frac{3\pi }{2})+1=-1+1=0$; $\cos (\frac{3\pi }{2})=0$. (Matches: $0=0$) $✓$

Final Answer:

The valid solutions are $x=0$ and $x=\frac{3\pi }{2}$.