Trigonometric Functions

Kindly consider the following answer

$\frac{\pi }{4}<x<\frac{\pi }{4}$

$\frac{14}{si{n}^{2}x}-2si{n}^{2}x=21-4+4si{n}^{2}x$

$t=\frac{-17\pm \sqrt[]{289+336}}{12}$

$=\frac{-17\pm 25}{12}=\frac{8}{12},\frac{-42}{12}$

$=\frac{2}{3},$

No. of solutions = 4

Kindly consider the following answer

$\frac{?}{4}<x<\frac{?}{4}$

$\frac{14}{si{n}^{2}x}?2si{n}^{2}x=21?4+4si{n}^{2}x$

$t=\frac{?17\pm \sqrt[]{289+336}}{12}$

$=\frac{?17\pm 25}{12}=\frac{8}{12},\frac{?42}{12}$

$=\frac{2}{3},$

No. of solutions = 4

$\frac{x^2-5x+6}{x^2-9}\ge -1\&\frac{x^2-5x+6}{x^2-9}\le 1$

$\frac{2x+1}{x+3}\ge 0,x\ne -3\&\frac{1}{x+3}\ge 0,x\ne -3\Rightarrow x>-3$

$x\in \left[-\frac{1}{2},\infty \right]...........(i)$

$x^2-3x+2>0and{x}^2-3x+1\ne 0$

(x – 2) (x – 1) > 0 and $x\ne \frac{3\pm \sqrt[]{5}}{2}$

$x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

From (i) and (ii)   $x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

${\left(ta{n}^{-1}x\right)}^3+{\left(co{t}^{-1}x\right)}^3=k{\pi }^3,x\in R$

$\Rightarrow \left(ta{n}^{-1}x+co{t}^{-1}x\right)\left({\left(ta{n}^{-1}x+co{t}^{-1}x\right)}^2-3ta{n}^{-1}xco{t}^{-1}x\right)=k{\pi }^3$

$\Rightarrow \frac{1}{3}\le k<\frac{7}{8}$

$cos\left(x+\frac{\pi }{3}\right)cos\left(\frac{\pi }{3}-x\right)=\frac{1}{4}co{s}^{2}2x$

$x=-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi$

$\therefore$ total number of solution = 7.

$\underset{n\to \infty }{lim}6tan\left\{{\displaystyle \sum _{r=1}^{n}ta{n}^{-1}}\left(\frac{1}{r^2+3r+3}\right)\right\}$

$=\underset{n\to \infty }{lim}6tan\left\{{\displaystyle \sum _{r=1}^{n}ta{n}^{-1}}\left(\frac{\left(r+2\right)-\left(r+1\right)}{1+\left(r+2\right)\left(r+1\right)}\right)\right\}$

$=\underset{n\to \infty }{lim}6tan\left\{\frac{r}{2}-ta{n}^{-1}\left(2\right)\right\}$

$6*\frac{1}{2}=3$

sin x = 1 – sin² x

sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

Let base = b

$tan60°=\frac{h}{b}$

$tan30°=\frac{h?20}{b}$