1. If $\cos x= \cos y$, then

$x = 2n\pi + y$ or

$x = 2n\pi - y$

for some $n \in \Z$

2. For any $n \in \Z$

$\cos (2n\pi+y) = \cos(2n\pi-y) = \cos y$

$\text{Prove} \cos (2n\pi + y) = \cos (2n\pi - y) = \cos y$

$\text{Proof:}$

$\because \cos (A+B) = \cos A \cos B - \sin A \sin B$

$\cos (2n \pi + y) = \cos 2n \pi \cos y - \sin 2n\pi \sin y=\cos y$

$\cos (2n \pi - y) = \cos 2n \pi \cos y + \sin 2n\pi \sin y=\cos y$ ,$n \in \Z$

$\Rightarrow \cos (2n\pi + y) = \cos (2n\pi - y) = \cos y$

$\cos x = \cos y$

$\Rightarrow \cos x - \cos y = 0$

$\cos A - \cos B = 2\sin \frac{(B-A)}{2} \sin \frac{(B+A)}{2}$

$\Rightarrow 2 \sin\frac{(y-x)}{2} \sin\frac{(y+x)}{2}=0$

either $\sin \frac{(y-x)}{2}=0,$

or $\sin\frac{(y+x)}{2}=0$

$\sin (\frac {y-x}{2})=0$

$\Rightarrow \sin (\frac {x-y}{2})=0$

($\sin \theta=0$ $\Rightarrow \theta=n \pi$ $,n \in \Z$)

$\frac{x-y}{2}=n\pi$ for some $n \in \Z$

$x=2n\pi-y$ for some $n \in \Z$

$\sin (\frac {x+y}{2})=0$

$\frac{x+y}{2}=n\pi$ for some $n \in \Z$

$x=2n\pi-y$ for some $n \in \Z$

As $\cos x = \cos y$, when

either $\sin \frac{(y-x)}{2}=0, \Rightarrow x=2n\pi+y$

or $\sin \frac{(y+x)}{2}=0, \Rightarrow x=2n\pi-y$

$\cos x = -\frac{1}{2}$

$=\cos (\frac{\pi}{2}+\frac{\pi}{6})$

$\cos x = \cos\frac{2\pi}{3}$

• If $\tan x = \tan y$, then $\\x = n\pi + y$ for some $n \in \Z$

• For any integer $n \in \Z \\ \tan (n \pi +y) = \tan (y)$

$\text{Prove} \tan (n \pi +y) = \tan y$ for all $n \in \Z$

$\text{Proof:}$

$\tan (A+B) = \frac{\tan A +\tan B}{1- \tan A \tan B}$

$\tan (n \pi + y) = \frac{ \tan n\pi + \tan y}{1- \tan n\pi \ \tan y}$

For any $n \in \Z$

$\tan n \pi= \frac{ \sin n\pi}{ \cos n\pi}$ = 0 ($\because \sin n\pi = 0)$

$\Rightarrow \tan (n \pi + y) = \frac{ \tan n\pi + \tan y}{1- \tan n\pi \ \tan y}= \tan y$

$\tan x = \tan y$

$\Rightarrow \tan x- \tan y =0$

$\Rightarrow \frac{\sin x}{\cos x} - \frac{\sin y}{\cos y} =0$

$\Rightarrow \frac{\sin x \cos y - \cos x \sin y}{ \cos x \cos y} =0$

$\Rightarrow \sin x \cos y - \cos x \sin y=0$

$\Rightarrow \sin (x-y) = 0$

$\Rightarrow (x-y) = n\pi$

$\Rightarrow x = n\pi + y \quad$ for some $n \in \Z$

Find the general solution to $a\cos \theta +b \sin \theta = c$

(where, a, b, c are real).

$\frac{a}{\sqrt{a^2 + b^2}} \cos \theta + \frac{b}{\sqrt{a^2 + b^2}} \sin \theta = \frac{c}{\sqrt{a^2 + b^2}}$

$\frac{a}{\sqrt{a^2 + b^2}} = \cos \phi$

$\frac{b}{\sqrt{a^2 + b^2}} = \sin \phi$

$\cos \phi \cos\theta + \sin \phi \sin \theta = \frac{c}{\sqrt{a^2+b^2}}$

$\cos (\theta- \phi) = \frac{c}{\sqrt{a^2+b^2}} = \cos y$

$\Biggl \vert \frac{c}{\sqrt{a^2+b^2}} \Biggl \vert \leq 1$

We know that for any $n \in \Z$

$\cos (2n\pi+y) = \cos (2n\pi-y) = cos y$

$\Rightarrow \theta - \phi = 2n \pi \pm y$ for all $n \in \Z$

$\theta \in \{\phi + 2n \pi \pm y \vert \ n \in \Z\}$

The solution set of the system of equations

$x+y = \frac{2 \pi}{3}$

$\cos x + \cos y = \frac{3}{2}$

where $x, y$ are real is __?

$y = \frac{2 \pi}{3} -x$

$\cos x + \cos \biggl(\frac{2 \pi}{3} -x\biggl) = \frac{3}{2}$

$\cos x + \cos \frac{2\pi}{3} \cos x + \sin \frac{2\pi}{3} \sin x = \frac{3}{2}$

$\cos x - \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \frac{3}{2}$

$\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \frac{3}{2}$

$\cos \frac{\pi}{3} \cos x + \sin \frac{\pi}{3} \sin x = \frac{3}{2}$

$\cos \biggl(x-\frac{\pi}{3} \biggl) = \frac{3}{2} \rightarrow$ no solution

Find the general value of $\theta$ satisfying

$2 \sin^2 \theta - 3 \sin \theta -2 = 0$

$z = \sin \theta$

$2z^2 - 3z-2 = 0$

$z = \frac {3 \pm\sqrt{9+16}}{4}$

$z = \frac{3 \pm 5}{4}, \quad z = \frac{3-5}{4} = \frac{-1}{2}$

$\sin \theta = \frac{-1}{2}$

$\sin \theta =-\frac{1}{2}$

$\because \sin (\pi+x) = - \sin x \Rightarrow x=\frac{\pi}{6}$

$\sin \frac{7\pi}{6}=-\sin \frac{\pi}{6} =-\frac{1}{2}$

$\therefore \sin \theta = \sin \frac{7\pi}{6}$

$\sin x = \sin y$

$x \in \{ n\pi + (-1)^ny | n \in \Z \}$

Therefore General solution is , $\{ n\pi + (-1)^n \frac{7\pi}{6} | n \in \Z \}$

Find the general value of $\theta$ satisfying

$\tan^2 \theta + \sec2\theta=1$

$\tan^2 \theta + \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta - \sin^2 \theta} =1$

$\tan^2 \theta + \frac{1+ \tan^2 \theta}{1 - \tan^2 \theta}=1$

$\tan^2 \theta (1-\tan^2 \theta) +1 +\tan^2 \theta = 1-\tan^2 \theta$

$2 \tan^2 \theta +1- \tan^4 \theta = 1-\tan^2 \theta$

$\tan^4 \theta - 3\tan^2 \theta = 0$

$\tan^2 \theta( \tan^2 \theta -3) = 0$

$\tan \theta=0 = \tan 0 \Rightarrow \theta=n \pi$

$\tan^2 \theta -3 =0$

$\Rightarrow \tan \theta=+\sqrt{3},\tan \theta=-\sqrt{3}$

$\tan \theta=\sqrt{3}=\tan \frac{\pi}{3}$

$\tan x=\tan y$

$x \in \{ n\pi +y | n \in \Z \}$

$\theta \in \{ n\pi +\frac{\pi}{3} | n \in \Z \}$

$\tan \theta = - \sqrt{3} = \tan (-\frac{\pi}{3})$

$\theta \in \{ n\pi -\frac{\pi}{3} | n \in \Z \}$

$\tan ^2 \theta+\sec^2 \theta=1 \$

$\Rightarrow \tan ^2 \theta (\tan ^2 \theta-3 )=0 \$

$\Rightarrow \tan \theta=0 \ or \ \tan \theta=\sqrt{3} \ or \ \tan \theta=-\sqrt{3}\$

Therefore General solution is ,$\{ n \pi | n \in \Z\} \cup \{n \pi+\frac{\pi}{3} \mid n \in \Z \} \cup \{n \pi-\frac{\pi}{3} \mid n \in \Z \}$

Find the values of $x \in(-\pi, \pi)$ which satisfy the equation

$2^{1+|\cos x|+ |\cos ^2 x |+ \cdots }=4$

$m \in \Z, \quad |\cos ^m x |=(|\cos x|)^m$

$1+|\cos x|+ |\cos ^2 x |+\cdots$

$= 1+|\cos x|+|\cos x|^2 +|\cos x|^3+ \cdots \cdots$

Put, $c = |\cos x|$

$= 1+c+c^2+c^3+\cdots$

$= \frac{1}{1-c}$ iff $|c|<1$

$x \neq 0$ and $x \in(-\pi, \pi)$

$\frac{1}{2^{1-|\cos x|}}=4$

$\Rightarrow 1-|\cos x|=\frac{1}{2}$

$\Rightarrow |\cos x|=\frac{1}{2}$

either $\cos x=\frac{1}{2}$ or $\cos x=-\frac{1}{2}$

$x \in \{2n\pi \pm \frac{\pi}{3} | n \in \Z \} \cup \{2n\pi \pm \frac{2\pi}{3}| n \in \Z\}$

So, values of x∈(−π,π) which satisfy the given equation are ,

$\frac{- \pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3},\frac{-2\pi}{3}$

[$2\pi - \frac{2 \pi}{3} =\frac{4\pi}{3} ∉ (−π,π)$]

Let m be an odd integer.

If $\sin mx = \sum_{p=0}^m b_p \sin^px$ for every $x$, then find the value of $b_0$ and $b_1.$

$\sin mx = b_0 + b_1\sin x + b_2 \sin^2x + …. +b_m\sin^m x$

Put x= 0 ,

$b_0 =0$

$\lim_{x \to 0} \frac{\sin mx}{\sin x} = \lim_{x \to 0}(b_1 + b_2 \sin x + b_3 \sin^2 x + …+b_m \sin^{m-1}x)$

$m = b_1$

$\therefore b_1=m_1, \quad b_0=0$

Find all the solution of

$4\cos^2 x \sin x - 2\sin^2 x = 3 \sin x$

$\sin x[4(1-\sin^2 x) - 2\sin x - 3] = 0$

$\sin x[1- 2\sin x - 4 \sin^2 x] =0$

For $\sin x = 0 ,x \in \{n\pi | n\in z \}$

For $4 \sin^2 x + 2\sin x -1 =0$

$\sin x = \frac{-2 \pm \sqrt{4+16}}{8}$

$=\biggl\{ \frac{-1 -\sqrt{5}}{4}, \frac{\sqrt{5} -1 }{4} \biggl\}$

$=\biggl\{ \sin\frac{-3\pi}{10}, \sin\frac{\pi}{10} \biggl\}$

$\sin x = \sin\frac{\pi}{10}$

$\Rightarrow x \in \{ n\pi + (-1)^n \frac{\pi}{10} | n \in \Z \}$

$\sin x = \sin \biggl(\frac{-3 \pi}{10}\biggl)$

$\Rightarrow x \in \{ n\pi + (-1)^n\frac{-3\pi}{10} \biggl\vert n \in \Z \}$

$\Rightarrow$ General solution is $\{n\pi | n\in \Z \} \cup \{n\pi + (-1)^n\frac{\pi}{10} | n\in \Z \}\cup \{n\pi + (-1)^n\frac{-3\pi}{10} | n\in \Z \}$

Thank you