$\sin x = \sin y$

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

$\cos x = \cos y$

$x \in \{ 2n \pi \pm y | n \in \Z\}$

$\tan x = \tan y$

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

$\text{Find the general solution to}$ $\cosec x = \cot x + √3$

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

$\text{We \ know \ that} \ 1-\cos x=2\sin^2 \frac{x}{2}$

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

$\quad =2 \biggl(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x \biggl)$

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

$=2 \sin \biggl(x + \frac{\pi}{6} \biggl)$

$\Rightarrow 2 \sin \biggl(x + \frac{\pi}{6} \biggl) = 1$

$\sin \biggl(x + \frac{\pi}{6} \biggl) = \frac{1}{2} = \sin \frac{\pi}{6}$

$x + \frac{\pi}{6} \in \biggl\{ n \pi + (-1)^n \frac{\pi}{6} | n \in \Z \biggl\}$

$x \in \biggl\{ n \pi + (-1)^n \frac{\pi}{6} - \frac{\pi}{6} | n \in \Z \biggl\}$

$\text{Find all solutions to}$ $\cosec \theta + \sec \theta =1$

$\frac{1}{\sin \theta} +\frac{1}{\cos \theta} =1$

$\cos \theta + \sin \theta = \sin \theta \cos \theta$

Let, $t\triangleq \sin \theta + \cos \theta, \\ \Rightarrow t^2 = \sin^2\theta + \cos^2 \theta + 2\sin \theta \cos \theta \\ \Rightarrow t^2 =1+2\sin \theta \cos \theta$

$\sin \theta \cos \theta = \frac{(t^2-1)}{2}$

$t = \frac{(t^2-1)}{2}$

$t^2-1 = 2t$

$t^2-2t-1=0$

$t = \frac{2 \pm \sqrt{8}}{2} = (1\pm\sqrt{2})$

$t = \sin \theta + \cos \theta = \sqrt{2} \biggl( \sin \theta \frac{1}{\sqrt{2}} + \cos \theta \frac{1}{\sqrt{2}}\biggl)$

$= \sqrt{2} \biggl( \sin \theta \cos\frac{\pi}{4} + \cos \theta \sin\frac{\pi}{4}\biggl)$

$= \sqrt{2} \sin \biggl(\theta + \frac{\pi}{4} \biggl)$

$|t| \leq \sqrt{2}$

$t = \sqrt{2} \sin(\theta + \frac{\pi}{4} ) = (1-\sqrt{2})$

$\sqrt{2} \sin \biggl(\theta + \frac{\pi}{4} \biggl) = \biggl(1-\sqrt{2}\biggl)$

$\sin\biggl(\theta + \frac{\pi}{4} \biggl) = \biggl(\frac{1}{\sqrt{2}}-1 \biggl) = \sin \phi$

$\Rightarrow \theta+\frac{\pi}{4} \in\biggl\{n \pi+(-1)^n \phi \mid n \in \Z \biggl\}$

$\Rightarrow \theta \in\biggl\{n \pi+(-1)^n \phi-\frac{\pi}{4} \mid n \in \Z \biggl\}$

$\text{Find the general solution to}$ $(1+ \sec 2 \theta)(1+ \sec 4\theta) = \cot \theta$

$\biggl(1+ \frac{1}{ \cos2 \theta}\biggl)\biggl(1+ \frac{1}{ \cos4 \theta}\biggl) = \frac{ \cos \theta}{ \sin \theta}$

$\text{Multipy L.H.S. and R.H.S. by}$ $\cos 2\theta \cos4\theta \sin\theta$

$(1+ \cos2 \theta)(1+ \cos4 \theta) \sin \theta = \cos \theta \cos2 \theta \cos4\theta$

$2 \cos^2 \theta \ 2 \cos^2 2\theta \sin\theta = \cos \theta \cos2 \theta \cos4 \theta$

$4 \sin \theta \cos^2\theta \cos^2 2 \theta - cos \theta \cos2 \theta \cos4 \theta =0$

$\cos \theta \cos2 \theta(4 \sin \theta \cos\theta \cos 2 \theta - \cos 4 \theta) =0$

$\cos \theta \cos2 \theta(2 \sin2 \theta \cos2 \theta - \cos 4 \theta) =0$

$\cos \theta \cos2 \theta ( \sin4 \theta - \cos4 \theta) = 0$

$\cos \theta =0, or \cos2 \theta =0, or \sin 4 \theta - \cos4 \theta =0$

$\cos \theta =0 \Leftrightarrow \theta \in \biggl\{ (2n \pm 1) \frac{\pi}{2} | n \in \Z \biggl\}$

$\cos2 \theta =0 \Leftrightarrow \theta \in \biggl\{ (2n \pm 1) \frac{\pi}{4} | n \in \Z \biggl\}$

$\sin 4 \theta - \cos 4 \theta =0$

$\Leftrightarrow \sin4 \theta \ \frac{1}{\sqrt{2}} - \cos4 \theta \ \frac{1}{\sqrt{2}}=0$

$\Leftrightarrow \sin4 \theta \cos\frac{\pi}{4} - \cos4 \theta \sin\frac{\pi}{4} =0$

$\Leftrightarrow \sin (4 \theta - \frac{\pi}{4}) = 0$

$\Leftrightarrow 4 \theta - \frac{\pi}{4} \in \{ n \pi | n \in \Z \}$

$\theta \in \{ \frac{n \pi}{4} + \frac{\pi}{16} | n \in \Z \}$

$(1 + \sec 2 \theta)(1 + \sec 4 \theta) = \cot \theta$

$\text{General solution}$

$\{ (2n \pm 1) \frac{\pi}{2} | n \in \Z \} \cup \{ (2n \pm 1) \frac{\pi}{4} | n \in \Z \} \cup \{ \frac{n \pi}{4} + \frac{\pi}{16} | n \in \Z\}$

$\text{Find the smallest positive value of x such that }$

$\tan(x + 100^{\circ}) = \tan(x + 50^{\circ}) \tan(x) \tan(x - 50^{\circ})$

$\frac{\sin(x + 100^{\circ})}{\cos(x + 100^{\circ})} = \frac{\sin(x + 50^{\circ}) \sin(x) \sin(x - 50^{\circ})}{\cos(x + 50^{\circ}) \cos(x) \cos(x - 50^{\circ})}$

Multiply L.H.S. and R.H.S. by $\cos(x+100) \cos(x+50) \cos(x) \cos(x-50)$

$\Rightarrow 4 \sin(x+100) \cos(x +50) \cos(x) \cos(x-50) = 4 \sin(x+50) \sin(x) \sin(x-50) \cos(x+100)$

L.H.S.

$(2 \sin(x+100) \cos x)(2 \cos(x+50) \cos(x-50))$

$= (\sin(2x+100) + \sin 100) \times (\cos 2x + \cos100)$

$= \sin(2x +100) \cos2x + \sin(2x +100) \cos100 + \sin100 \cos2x + \sin100 \cos100$

R.H.S.

$4 \sin(x+50) \sin x \sin(x-50) \cos(x+100)$

$=(2 \sin(x+50) \sin(x-50) \times (2 \sin x \cos(x+100))$

$=(\cos100 - \cos2x) \times(\sin(2x+100) -\sin100)$

$=\sin(2x+100)\cos100 - \sin100 \cos100 -\cos2x \sin(2x+100) + \cos2x \sin100$

$\because$ L.H.S. = R.H.S.

$\Rightarrow \sin(2x+100) \cos2x + \sin(2x+100) \cos100 + \sin100 \cos2x + \sin100 \cos100 \quad =\sin(2x+100)\cos100 - \sin100 \cos100 -\cos2x \sin(2x+100) + \cos2x \sin100$

$2 \sin(2x+100) \cos2x + 2 \sin100 \cos100 = 0$

$2 \sin(2x + 100) \cos2x + 2 \sin100 \cos100 = 0$

$\sin(4x + 100) + \sin100 + \sin200 =0$

$\sin(4x + 100) + 2 \sin150 \cos50 = 0$ ($\because \sin 150 = 1/2)$

$\sin(4x + 100) = -\cos50 = -\sin40$

$4x + 100 \in \{n \pi+(-1)^n[(-40) \times \frac{\pi}{180}] |n \in \Z \}$

$4x \in \{ n \pi + (-1)^n [\frac{-2 \pi}{9}] \frac{-5 \pi}{9} | n \in \Z\}$

$x \in \{ \frac{n\pi}{4} +(-1)^n[\frac{-\pi}{18}] -\frac{5\pi}{36} | n \in \Z\}$

$n=1, \quad \frac{\pi}{4} - \biggl( \frac{-\pi}{18} \biggl) - \frac{5\pi}{36}$

$=\frac{\pi}{4} + \frac{\pi}{18} - \frac{5\pi}{36}$

$=45^{\circ} + 10^{\circ} - 25^{\circ}$

$=30^{\circ}$

$=\frac{\pi}{6}$

$\text {For all}$ $x \in \biggl[0, \frac{\pi}{2}\biggl]$, $\text{ show that}$ $\cos (\sin x) \geq \sin (\cos x)$

$\sin \biggl(\frac{\pi}{2} - \sin x\biggl) \geq \sin (\cos x)$

Let , $A =\biggl(\frac{\pi}{2} - \sin x\biggl) \quad\quad B=(\cos x), x\in \biggl[{0,\frac{\pi}{2}}\biggl]$

$\frac{\pi}{2} -1 \leq \frac{\pi}{2} - \sin x \leq \frac{\pi}{2}$

$0 \leq \cos x \leq 1$

$\biggl[\frac{\pi}{2} -1, \frac{\pi}{2}\biggl] \subset \biggl[0,\frac{\pi}{2} \biggl]$

$[0,1] \subset \biggl[0, \frac{\pi}{2}\biggl]$ $\Rightarrow A,B \in \biggl[0, \frac{\pi}{2}\biggl]$

Now we know that,

Here, A $\geq$ B $\Leftrightarrow$ $\sin A \geq \sin B$

Showing $\frac{\pi}{2} - \sin x \geq \cos x \ in \ x \in [0, \frac{\pi}{2}]$ is equivalent to showing

$\cos x + \sin x \leq \frac{\pi}{2} \ for \ all \ x \in [0,\frac{\pi}{2}].$

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

=$\sqrt{2}\sin \biggl(x+\frac{\pi}{4}\biggl) ,\quad\quad x \in \biggl[0, \frac{\pi}{2} \biggl]$

$\sin \biggl(x+\frac{\pi}{4}\biggl) \leq 1$

$\therefore \sin x + \cos x \leq \sqrt{2}$

$\because \sqrt{2} < \frac{\pi}{2}$

($\because 1.41 < 1.57$)

$\Rightarrow \sin x + \cos x <\frac{\pi}{2} \ for \ all \ x \in \biggl[0, \frac{\pi}{2}\biggl]$

$\Rightarrow \frac{\pi}{2} - \sin x \geq \cos x \ for \ all \ x \in \biggl[0, \frac{\pi}{2}\biggl]$

$\Rightarrow \sin \biggl(\frac{\pi}{2} - \sin x\biggl) \geq \sin (\cos x)$ $\Rightarrow \cos (\sin x) \geq \sin (\cos x)$

$\text{Find the smallest positive number p for which the equation}$

$\cos (p \sin x) = \sin(p \cos x)$ $\text{has a solution}$ where, $x\in [0, 2\pi]$

$\sin(\frac{\pi}{2} - p \sin x) = \sin(p \cos x)$

$\frac{\pi}{2} - p \sin x = n \pi + (-1)^n p \cos x \ for \ some \ n \in \Z$

For $n=0$, $\frac{\pi}{2} - p \sin x = p \cos x$

$p( \sin x + \cos x ) = \frac{\pi}{2}$

$p\sqrt{2} \sin (x + \frac{\pi}{4}) = \frac{\pi}{2}$

$\sin (x + \frac{\pi}{4}) \leq 1$

$\Rightarrow p \geq \frac{n}{2\sqrt{2}}$

For $n=1$

$\frac{\pi}{2} - p \sin x = \pi - p \cos x$

$p\sqrt{2} . \cos (x + \frac{\pi}{4}) = \frac{\pi}{2}$

Similarly here , $p \geq \frac{n}{2\sqrt{2}}$

$\frac{\pi}{2} - p \sin x = n \pi + (-1)^n p \cos x$

$p((-1)^n \cos x + \sin x) = \frac{-\pi}{2}(2n-1)$

$p\sqrt{2}(\frac{(-1)^n}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) = -\frac{\pi}{2}(2n-1)$

$\because |a \cos x + b \sin x | \leq \sqrt(a^2+b^2)$

$\Rightarrow |p|\sqrt{2} \geq \frac{\pi}{2}|2n-1|$

$p \geq \frac{\pi}{2\sqrt{2}}$