Solve the following equations:

(i) Given:

$\sin x+\cos x=\sqrt{2} \ldots$.(i)

The equation is of the form $a \sin x+b \cos x=c$, where $a=1, b=1$ and $c=\sqrt{2}$.

Let: $a=r \sin \alpha$ and $b=r \cos \alpha$

Now, $r=\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$ and $\tan \alpha=1 \Rightarrow \alpha=\frac{\pi}{4}$

On putting $a=1=r \sin \alpha$ and $b=1=r \cos \alpha$ in equation (i), we get:

$r \sin \alpha \sin x+r \cos \alpha \cos x=\sqrt{2}$

$\Rightarrow r \cos (x-\alpha)=\sqrt{2}$

$\Rightarrow \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=\sqrt{2}$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=1$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\cos 0$

$\Rightarrow x-\frac{\pi}{4}=n \pi \pm 0, n \in \mathrm{Z}$

$\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}, n \in \mathrm{Z}$

$\Rightarrow x=(8 n+1) \frac{\pi}{4}, n \in Z$

(ii) Given: $\sqrt{3} \cos x+\sin x=1 \ldots$ (ii)

The equation is of the form of $a \cos x+b \sin x=c$, where $a=\sqrt{3}, b=1$ and $c=1$.

Let: $a=r \cos \alpha$ and $b=r \sin \alpha$

Now, $r=\sqrt{a^{2}+b^{2}}=\sqrt{(\sqrt{3})^{2}+1^{2}}=2$ and $\tan \alpha=\frac{b}{a}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\frac{\pi}{6}$

On putting $a=\sqrt{3}=r \cos \alpha$ and $b=1=r \sin \alpha$ in equation (ii), we get:

$r \cos \alpha \cos x+r \sin \alpha \sin x=1$

$\Rightarrow r \cos (x-\alpha)=1$

$\Rightarrow 2 \cos (x-\alpha)=1$

$\Rightarrow \cos \left(x-\frac{\pi}{6}\right)=\frac{1}{2}$

$\Rightarrow \cos \left(x-\frac{\pi}{6}\right)=\cos \frac{\pi}{3}$

$\Rightarrow x-\frac{\pi}{6}=2 n \pi \pm \frac{\pi}{3}, \mathrm{n} \in \mathrm{Z}$

On taking positive sign, we get:

$x-\frac{\pi}{6}=2 n \pi+\frac{\pi}{3}$

$\Rightarrow \mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{3}+\frac{\pi}{6}$

$\Rightarrow \mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$

$\Rightarrow x=(4 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$

Now, on taking negative sign of the equation, we get:

$x-\frac{\pi}{6}=2 m \pi-\frac{\pi}{3}, \mathrm{~m} \in \mathrm{Z}$

$\Rightarrow \mathrm{x}=2 \mathrm{~m} \pi-\frac{\pi}{3}+\frac{\pi}{6}, \mathrm{~m} \in \mathrm{Z}$

$\Rightarrow \mathrm{x}=2 \mathrm{~m} \pi-\frac{\pi}{6}=(12 \mathrm{~m}-1) \frac{\pi}{6}, \mathrm{~m} \in \mathrm{Z}$

(iii) Given: $\sin x+\cos x=1 \quad$...(iii)

The equation is of the form $a \sin \theta+b \cos \theta=c$, where $a=1, b=1$ and $c=1$.

Let: $a=r \sin \alpha$ and $b=r \cos \alpha$

Now, $r=\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$ and $\tan \alpha=\frac{b}{a}=1 \Rightarrow \alpha=\frac{\pi}{4}$

On putting $a=1=r \sin \alpha$ and $b=1=r \cos \alpha$ in equation (iii), we get:

$r \sin \alpha \sin x+r \cos \alpha \cos x=1$

$\Rightarrow r \cos (x-\alpha)=1$

$\Rightarrow \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=1$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}$

$\Rightarrow x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4}, n \in Z$

On taking positive sign, we get:

$x-\frac{\pi}{4}=2 n \pi+\frac{\pi}{4}$

$\Rightarrow \mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{4}+\frac{\pi}{4}$

$\Rightarrow \mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{2}, n \in Z$

On taking negative sign, we get:

$x-\frac{\pi}{4}=2 m \pi-\frac{\pi}{4}$

$\Rightarrow \mathrm{x}=2 m \pi, m \in \mathrm{Z}$

(iv) Given: $\operatorname{cosec} x=1+\cot x$

$\Rightarrow \frac{1}{\sin x}=1+\frac{\cos x}{\sin x}$

$\Rightarrow \sin x+\cos x=1$     .....(iv)

The equation is of the form $a \sin x+b \cos x=c$, where $a=1, b=1$ and $c=1$.

Let: $a=r \sin \alpha$ and $b=r \cos \alpha$

Now, $r=\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$ and $\tan \alpha=1 \Rightarrow \alpha=\frac{\pi}{4}$

On putting $a=1=r \sin \alpha$ and $b=1=r \cos \alpha$ in equation (iv), we get:

$r \sin \alpha \sin x+r \cos \alpha \cos x=1$

$\Rightarrow r \cos (x-\alpha)=1$

$\Rightarrow \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=1$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}$

$\Rightarrow x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$

On taking positive sign, we get:

$x=2 n \pi+\frac{\pi}{2}, n \in Z$

On taking negative sign, we get:

$x=2 m \pi, m \in Z$