Question

Solve the following equations:
(i) $\sin \mathrm{x}+\cos \mathrm{x}=\sqrt{2}$
(ii) $\sqrt{3}\cos x+\sin x=1$
(iii) $\sin x+\cos x=1$
(iv) $\mathrm{cosec}x=1+\cot x$

Answer 1

(i) Given:
$\sin x+\cos x=\sqrt{2}$ ...(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 $\tan \alpha =1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$in equation (i), we get:
$r\sin \alpha \sin x+r\cos \alpha \cos x=\sqrt{2}$
$$\begin{gathered} \Rightarrow r\cos (x-\alpha )=\sqrt{2} \\ \Rightarrow \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\sqrt{2} \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\cos 0 \\ \Rightarrow x-\frac{\mathrm{\pi }}{4}=n\mathrm{\pi }\pm 0,n\in \mathrm{Z} \\ \Rightarrow \mathrm{x}=\mathrm{n\pi }+\frac{\mathrm{\pi }}{4},\mathit{}n\in \mathrm{Z} \\ \Rightarrow x=(8n+1)\frac{\mathrm{\pi }}{4},n\in Z \end{gathered}$$

(ii) Given: $\sqrt{3}\cos x+\sin x=1$ ...(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 $\tan \alpha =\frac{b}{a}=\frac{1}{\sqrt{3}}\Rightarrow \alpha =\frac{\mathrm{\pi }}{6}$
On putting $a=\sqrt{3}=r\cos \alpha$ and $b=1=r\sin \alpha$ in equation (ii), we get:
$r\cos \alpha \cos x+r\sin \alpha \sin x=1$

$$\begin{gathered} \Rightarrow r\cos (x-\alpha )\hspace{0.17em}=1 \\ \Rightarrow 2\cos (x-\alpha )=1 \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2} \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{6}\right)=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow x-\frac{\mathrm{\pi }}{6}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z} \end{gathered}$$

On taking positive sign, we get:
$$\begin{gathered} x-\frac{\mathrm{\pi }}{6}=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{3} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z} \\ \Rightarrow x=(4\mathrm{n}+1)\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z} \end{gathered}$$
Now, on taking negative sign of the equation, we get:
$$\begin{gathered} x-\frac{\mathrm{\pi }}{6}=2m\mathrm{\pi }-\frac{\mathrm{\pi }}{3},\mathrm{m}\in \mathrm{Z} \\ \Rightarrow \mathrm{x}=2\mathrm{m\pi }-\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6},\mathrm{m}\in \mathrm{Z} \\ \Rightarrow \mathrm{x}=2\mathrm{m\pi }-\frac{\mathrm{\pi }}{6}=(12\mathrm{m}-1)\frac{\mathrm{\pi }}{6},\mathrm{m}\in \mathrm{Z} \end{gathered}$$

(iii) Given: $\sin x+\cos x=1$ ...(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 $\tan \alpha =\frac{b}{a}=1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$ in equation (iii), we get:
$r\sin \alpha \sin x+r\cos \alpha \cos x=1$

$$\begin{gathered} \Rightarrow r\cos (x-\alpha )=1 \\ \Rightarrow \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},n\in Z \end{gathered}$$

On taking positive sign, we get:
$$\begin{gathered} x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2},\mathit{}n\in \mathit{}Z \end{gathered}$$
On taking negative sign, we get:
$$\begin{gathered} x-\frac{\mathrm{\pi }}{4}=2m\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=2m\mathrm{\pi },m\mathit{}\in \mathrm{Z} \end{gathered}$$

(iv) Given: $\cos ecx=1+cotx$
$$\begin{gathered} \Rightarrow \frac{1}{\sin x}=1+\frac{\cos x}{\sin x} \\ \Rightarrow \sin x+\cos x=1...\left(\mathrm{iv}\right) \end{gathered}$$

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 $\tan \alpha =1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$ in equation (iv), we get:

$$\begin{gathered} r\sin \alpha \sin x+r\cos \alpha \cos x=1 \\ \Rightarrow r\cos (x-\alpha )=1 \\ \Rightarrow \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{Z} \end{gathered}$$
On taking positive sign, we get:
$x=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2},n\in Z$
On taking negative sign, we get:
$x=2m\pi ,m\in Z$