AP EAPCET 2025 21st May Morning Shift Question Paper with Solutions

$\begin{array}{cc}& \text{We have }𝑓\left(𝑥\right)=\cos 𝑥-3\\ & \begin{array}{cc}& \text{ Domain }=𝑅\\ & \text{ Range }-1\le \cos 𝑥\le 1\\ & -1-3\le \cos 𝑥-3\le 1-3\\ & -4\le 𝑓\left(𝑥\right)\le -2\\ & \text{ Range }=[-4,-2]\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }𝑓\left(𝑥\right)=2𝑥-3\text{ and }𝑔\left(𝑥\right)=5{𝑥}^2-2\\ & 𝑔\circ 𝑓\left(𝑥\right)=5\left[𝑓\right(𝑥){]}^2-2=5(2𝑥-3{)}^2-2\\ & \because \text{ Least value of }𝑔\circ 𝑓\left(𝑥\right)=-2\end{array}$

$\begin{array}{cc}& \text{ We have, }{1}^3+2^3+3^3+\dots +{𝑛}^3>𝑥\\ & \because \Sigma {𝑛}^3={\left[\frac{𝑛(𝑛+1)}{2}\right]}^2=\frac{{𝑛}^2}{4}(𝑛+1{)}^2>𝑥\\ & \because \frac{{𝑛}^2}{4}<\frac{{𝑛}^2}{4}(𝑛+1{)}^2\\ & \therefore 𝑥=\frac{{𝑛}^2}{4}\end{array}$

(i) $𝐴𝐵=0\Rightarrow 𝐴=0$ or $𝐵=0$

Which is false.

(ii) If $𝐴𝐵=𝐵𝐴={𝐼}_3\Rightarrow {𝐴}^{-1}=𝐵$

$\because$ It is true.

$\begin{array}{cc}\left(𝑖𝑖𝑖\right)(𝐴-𝐵{)}^2& =(𝐴-𝐵)(𝐴-𝐵)\\ & ={𝐴}^2-𝐴𝐵-𝐵𝐴+{𝐵}^2\end{array}$

$\Rightarrow {𝐴}^2-2𝐴𝐵+{𝐵}^2$ is true only when

$𝐴𝐵=𝐵𝐴$

$\because$ It is false.

Given, $𝐴=⎡⎣⎢\begin{array}{ccc}1& -1& 2\\ -2& 3& -3\\ 4& -4& 5\end{array}⎤⎦⎥$ then

${𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}1& -2& 4\\ -1& 3& -4\\ 2& -3& 5\end{array}⎤⎦⎥$

Now, $𝐴{𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}6& -11& 18\\ -11& 22& -35\\ 18& -35& 57\end{array}⎤⎦⎥$

Then, $𝐴{𝐴}^{𝑇}-𝐴-{𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}4& -8& 12\\ -8& 16& -28\\ 12& -28& 47\end{array}⎤⎦⎥$

We have, $𝐴=⎡⎣⎢\begin{array}{ccc}𝑥& 2& 1\\ -2& 𝑦& 0\\ 2& 0& -1\end{array}⎤⎦⎥$

$\begin{array}{cc}𝑡𝑟𝑎𝑐𝑒𝑜𝑓\left(𝐴\right)& =𝑥+𝑦-1=0...\left(𝑖\right)\\ \det 𝐴& =-2𝑦-𝑥𝑦-4=-6\\ 2𝑦+𝑥𝑦& =2...\left(𝑖𝑖\right)\end{array}$

Solving Eqs. (i) and (ii), we get

$\begin{array}{cc}& 2𝑦+𝑦(1-𝑦)=2\\ \Rightarrow & {𝑦}^2-3𝑦+2=0\end{array}$

$𝑦=1\text{ and }2$

and $𝑥=0$ and -1

Then, the minor of the element 1 is

$\begin{array}{cc}& \mid \begin{array}{cc}-2& 𝑦\\ 2& 0\end{array}\mid \\ & =-2𝑦=-2\left(2\right)=-4\end{array}$

$\begin{array}{cc}& \text{ We have, }𝑖=\sqrt{\surd -1}\\ & \begin{array}{cc}& \sum _{𝑛=2}^{30}{𝑖}^{𝑛}={𝑖}^2+{𝑖}^3+\dots +{𝑖}^{30}=0+{𝑖}^{30}=0-1\\ & \sum _{𝑛=30}^{65}{𝑖}^{𝑛+3}={𝑖}^{33}+{𝑖}^{34}+\dots +{𝑖}^{68}=0\\ & \because \sum _{𝑛=2}^{30}{𝑖}^{𝑛}+\sum _{𝑛=30}^{65}{𝑖}^{𝑛+3}=-1+0=-1\end{array}\end{array}$

${𝑧}_1$ and ${𝑧}_2$ are both $𝑛$th roots of unity. This means each one can be written in the form ${𝑒}^{\frac{𝑖2𝜋𝑟}{𝑛}}$, where $𝑟$ is an integer.

Let ${𝑧}_1={𝑒}^{\frac{𝑖2𝜋{𝑟}_1}{𝑛}}$ and ${𝑧}_2={𝑒}^{\frac{𝑖2𝜋{𝑟}_2}{𝑛}}$ for some integers ${𝑟}_1$ and ${𝑟}_2$.

If the line segment joining ${𝑧}_1$ and ${𝑧}_2$ makes a right angle at the origin, this means the points ${𝑧}_1$, ${𝑧}_2$, and the origin form a right-angled triangle with the right angle at the origin.

This can only happen if $\frac{{𝑧}_1}{{𝑧}_2}$ is a purely imaginary number. A complex number is purely imaginary when its argument (angle) is $\frac{𝜋}{2}$ or $\frac{3𝜋}{2}$, or in general, $\frac{𝜋}{2}+𝑚𝜋$ where $𝑚$ is an integer.

We find: $\frac{{𝑧}_1}{{𝑧}_2}={𝑒}^{\frac{𝑖2𝜋}{𝑛}({𝑟}_1-{𝑟}_2)}$

For this to be purely imaginary, we need: $\frac{2𝜋}{𝑛}({𝑟}_1-{𝑟}_2)=\frac{𝜋}{2}+𝑚𝜋$

Divide both sides by $𝜋$: $\frac{2}{𝑛}({𝑟}_1-{𝑟}_2)=\frac{2𝑚+1}{2}$

Now solve for $𝑛$: $𝑛=\frac{4({𝑟}_1-{𝑟}_2)}{2𝑚+1}$

Since $𝑛$ must be a positive integer, $2𝑚+1$ must divide evenly into $4({𝑟}_1-{𝑟}_2)$. The simplest way for this to always work for integer values is when $𝑛$ is a multiple of $4$. Therefore, $𝑛=4𝑘$ for some positive integer $𝑘$.

Let, $𝑍=\sqrt{\surd \sqrt{\surd 2}+1}+𝑖\sqrt{\surd \sqrt{\surd 2}-1}$

$\begin{array}{cc}& \Rightarrow \left|𝑍\right|=\sqrt{\surd \sqrt{\surd 2}+1+\sqrt{\surd 2}-1}=\sqrt{\surd 2\sqrt{\surd 2}}\\ & \cos 𝜃=\frac{\sqrt{\surd \sqrt{\surd 2}+1}}{\sqrt{\surd 2\sqrt{\surd 2}}},\sin 𝜃=\frac{\sqrt{\surd \sqrt{\surd 2}-1}}{\sqrt{\surd 2\sqrt{\surd 2}}}\\ & \therefore {\cos }^2𝜃=\frac{\sqrt{\surd 2}+1}{2\sqrt{\surd 2}},{\sin }^2𝜃=\frac{\sqrt{\surd 2}-1}{2\sqrt{\surd 2}}\\ & \Rightarrow \frac{1+\cos 2𝜃}{2}=\frac{\sqrt{\surd 2}+1}{2\sqrt{\surd 2}}\\ & \Rightarrow \cos 2𝜃=\frac{1}{\sqrt{\surd 2}}\end{array}$

$\Rightarrow 2𝜃=\frac{𝜋}{4}$ (since, real and imaginary part are positive.)

Now,

${𝑍}^8=(\sqrt{\surd 2\sqrt{\surd 2}}{)}^8(\cos (8\times \frac{𝜋}{8})+𝑖\sin (8\times \frac{𝜋}{8}))$

$\begin{array}{cc}& =\left(2\sqrt{\surd 2}{)}^4\right(\cos 𝜋+𝑖\sin 𝜋)\\ & =(2^{\frac{3}{2}}\times 4)(-1+0)\\ & =2^6(-1)=-64\end{array}$

$\sqrt{\surd 2}{𝑥}^2-𝑏𝑥+(8-2\sqrt{\surd 5})=0$

Let it's root's be $𝛼$ and $𝛽$,

$𝛼+𝛽=\frac{𝑏}{\sqrt{\surd 2}},𝛼𝛽=\frac{8-2\sqrt{\surd 5}}{\sqrt{\surd 2}}$

The harmonic mean of the roots,

$\begin{array}{c}𝐻=\frac{2𝛼𝛽}{𝛼+𝛽}=4\\ 𝛼𝛽=2(𝛼+𝛽)\\ \frac{8-2\sqrt{\surd 5}}{\sqrt{\surd 2}}=2\cdot \left(\frac{𝑏}{\sqrt{\surd 2}}\right)\\ 2𝑏=2(4-\sqrt{\surd 5})\\ 𝑏=4-\sqrt{\surd 5}\end{array}$

We have,

$\begin{array}{cc}2𝑘{𝑥}^2-(4𝑘+1)𝑥+2& <0\\ \because 2𝑘{𝑥}^2-4𝑘𝑥-𝑥+2& <0\\ 2𝑘𝑥(𝑥-2)-1(𝑥-2)& <0\\ (𝑥-2)(2𝑘𝑥-1)& <0\\ 2<𝑥& <\frac{1}{2𝑘}\end{array}$

Expression is negative for exactly 3 integral values of $𝑥$ which is $𝑥=3,4,5$

$\begin{array}{cc}& 5<𝑥\le 6\\ \because & \frac{1}{2𝑘}\le 6\Rightarrow 𝑘\ge \frac{1}{12}\\ \therefore & \frac{1}{2𝑘}>5\Rightarrow 𝑘<\frac{1}{10}\\ \therefore & 𝑘\in [\frac{1}{12},\frac{1}{10})\end{array}$

We have, $𝑓\left(𝑥\right)=4{𝑥}^4+4{𝑥}^3-23{𝑥}^2$$-12𝑥+36=0$

Using concept a polynomial $𝑓\left(𝑥\right)$ has a multiple root $𝑟$, then $𝑓\left(𝑟\right)=0$ and

$\begin{array}{cc}& {𝑓}^{\prime }\left(𝑟\right)=0\\ & {𝑓}^{\prime }\left(𝑥\right)=16{𝑥}^3+12{𝑥}^2-46𝑥-12\end{array}$

Now, using hit and trial

By taking $𝑥=-2$

$\begin{array}{c}𝑓(-2)=4(-2{)}^4+4(-2{)}^3-23(-2{)}^2\\ -12(-2)+36\\ =64-32-92+24+36=0\\ \Rightarrow {𝑓}^{\prime }(-2)=16(-2{)}^3+12(-2{)}^2\\ -46(-2)-12\\ =-128+48+92-12=0\end{array}$

$\because 𝑥=-2$ is a multiple root.

By dividing the equation by $(𝑥+2{)}^2$, we get another multiple root $𝑥=\frac{3}{2}$

$\begin{array}{cc}& \text{ Given, }𝛼>𝛽\\ & \therefore 𝛼=\frac{3}{2}\text{ and }𝛽=-2\\ & \therefore 2𝛼-𝛽=3+2=5\end{array}$

Given, ${𝑥}^3-13{𝑥}^2+𝑘𝑥+189=0$

$\begin{array}{cc}& 𝛼+𝛽+𝛾=13,𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑘\\ & \text{ and }𝛼𝛽𝛾=-189\text{ and }𝛽-𝛾=2\\ & [\because 𝛾=𝛽-2]\\ & 𝛼+𝛽+𝛽-2=13\\ & \Rightarrow (15-2𝛽)𝛽(𝛽-2)=-189\\ & \Rightarrow 𝛽(15𝛽-2{𝛽}^2-30+4𝛽)=-189\\ & \Rightarrow -2{𝛽}^3+19{𝛽}^2-30𝛽+189=0\\ & \Rightarrow 2{𝛽}^3-19{𝛽}^2+30𝛽-189=0\\ & \Rightarrow (𝛽-9)(2{𝛽}^2-𝛽+21)=0\\ & \Rightarrow 𝛽=9,2{𝛽}^2-𝛽+21\ne 0\forall 𝛽\in 𝑅\\ & \therefore 𝛽=9,𝛾=7\end{array}$

and $𝛼=-3$

$𝑘=𝛼𝛽+𝛽𝛾+𝛾𝛼=-27+63-21=15$

Now, $\frac{𝛽+𝛾}{𝑘+𝛼}=\frac{9+7}{15-3}=\frac{16}{12}=\frac{4}{3}$

$\therefore 𝛽+𝛾:𝑘+𝛼=4:3$

We have, $𝑥𝑦𝑧=30$

$30=1,2,3,5,6,10,15,30$ are the integral factor of 30

$\begin{array}{cc}& 30=3\times 2\times 5=3!\\ & 30=6\times 5\times 1=3!\\ & 30=10\times 3\times 1=3!\\ & 30=15\times 2\times 1=3!\\ & 30=30\times 1\times 1=\frac{3!}{2!}\end{array}$

$\begin{array}{cc}\therefore \text{ Total number of factor }& =4\times 3!+\frac{3!}{2!}\\ & =24+3=27\end{array}$

Given word is INEONVENIENCE. The distinet letters are I, N, C, O, V, E (6 letters) and their frequencies are

$\begin{array}{cc}& I-2\\ & N-4\\ & C-2\\ & O-1\\ & V-1\\ & E-3\end{array}$

Total number of 5-letters words

(i) When 5 distinct letters, number of words $=6{𝑃}_5=720$

(ii) One pair and 3 distinct letters,

$\begin{array}{cc}\text{ number of words }& ={}^4{𝐶}_1\times {}^5{𝐶}_3\times \frac{5!}{2!}\\ & =4\times 10\times 60=2400\end{array}$

(iii) One triple and 2 distinct letters, number of words

$={}^2{𝐶}_1\times {}^5{𝐶}_2\times \frac{5!}{3!}$

$=2\times 10\times 20=400$

(iv) Two pairs and one distinct, number

$\begin{array}{cc}\text{ of words }& ={}^4{𝐶}_2\times {}^4{𝐶}_1\times \frac{5!}{2!2!}\\ & =6\times 4\times 30=720\end{array}$

(v) One triplet and one pair, number of words

$=4\times \frac{5!}{3!2!}=4\times 10=40$

(vi) One 4 a like and one different

$={}^1{𝐶}_1\times {}^5{𝐶}_1\times \frac{5!}{4!}=25$

$\therefore$ Number of words with at least one repealed letters

$\begin{array}{cc}& =(720+2400+400+720+40+25)-720\\ & =3585\end{array}$

Given, PERFECTION

Vowels E E I O and consonants PRFC T M.

Vowels can be arranged by $\frac{4!}{2!}$

$\begin{array}{cc}& =4\cdot 3!=2!3!\\ & 𝐸\times 𝑋𝐸\times 𝑋\times 𝑋𝑂\end{array}$

Consonants are arranged at $\times$ places by 6 ! ways.

$\because$ Total number of ways $=6!\times 3!\times 2!$

The given sum can be rewritten as

$\sum _{𝑛}^{𝑘=0}\sum _{𝑘}^{𝑟=0}{𝐶}_{𝑟}$

Now, changing order of summarion

$\sum _{𝑛}^{𝑟=0}\sum _{𝑛}^{𝑘=𝑟}{𝐶}_{𝑟}[\because \sum _{𝑛}^{𝑟=0}{𝐶}_{𝑟}=2^{𝑛}]$

$\begin{array}{cc}& \sum _{𝑟=0}^{𝑛}{𝐶}_{𝑟}\sum _{𝑘=𝑟}^{𝑛}1=\sum _{𝑟=0}^{𝑛}{𝐶}_{𝑟}(𝑛-𝑟+1)\\ & \Rightarrow (𝑛+1)\sum _{𝑟=0}^{𝑛}{𝐶}_{𝑟}-\sum _{𝑟=0}^{𝑛}𝑟\cdot {𝐶}_{𝑟}\\ & =(𝑛+1)2^{𝑛}-\sum _{𝑟=0}^{𝑛}𝑟\cdot {}^{𝑛-1}{𝐶}_{𝑟-1}\times \frac{𝑛}{𝑟}\\ & =(𝑛+1)2^{𝑛}-𝑛\sum _{𝑟=0}^{𝑛}𝑛-1{𝐶}_{𝑟}\\ & =(𝑛+1)2^{𝑛}-𝑛\cdot 2^{𝑛-1}\\ & =𝑛\cdot 2^{𝑛}+2^{𝑛}-𝑛\cdot 2^{𝑛-1}\\ & =2^{𝑛-1}(𝑛+2)\end{array}$

$\begin{array}{cc}& \text{ We have, }{\left(\frac{3𝑥-5}{4{𝑥}^2+3}\right)}^{-\frac{4}{5}}={\left(\frac{4{𝑥}^2+3}{3𝑥-5}\right)}^{\frac{4}{5}}\\ & \begin{array}{cc}& ={\left\{\frac{4{𝑥}^2(1+\frac{3}{4{𝑥}^2})}{3𝑥(1-\frac{5}{3𝑥})}\right\}}^{\frac{4}{5}}={\left(\frac{4𝑥}{3}\right)}^{\frac{4}{5}}\frac{{(1+\frac{3}{4{𝑥}^2})}^{4/5}}{{(1-\frac{5}{3𝑥})}^{4/5}}\\ & \Rightarrow {\left(\frac{4𝑥}{3}\right)}^{\frac{4}{5}}{(1+\frac{3}{4{𝑥}^2})}^{\frac{4}{5}}\cdot {(1-\frac{5}{3𝑥})}^{-\frac{4}{5}}\end{array}\end{array}$

$\begin{array}{cc}& ={\left(\frac{4𝑥}{3}\right)}^{\frac{4}{5}}(1+\frac{4}{5}\cdot \frac{3}{4{𝑥}^2}+\dots )\\ & {(1+\frac{5}{3𝑥}\cdot \frac{4}{5}+\frac{4}{5}(+\frac{9}{5})\left(\frac{5}{3𝑥}\right))}^2\frac{1}{2!}+\dots \dots \\ & ={\left(\frac{4𝑥}{3}\right)}^{\frac{4}{5}}(1+\frac{3}{5{𝑥}^2}+\dots )(1+\frac{4}{3𝑥}+\frac{2}{{𝑥}^2})\\ & ={\left(\frac{4𝑥}{3}\right)}^{\frac{4}{5}}(1+\frac{4}{3𝑥}+{(\frac{3}{5}+2)}^{\frac{1}{{𝑥}^2}})\\ & ={\left(\frac{4𝑥}{3}\right)}^{\frac{4}{5}}(1+\frac{4}{3𝑥}+\frac{13}{5{𝑥}^2})\end{array}$

$\begin{array}{cc}& \text{ Given, }𝐻\left(𝑥\right)=3{𝑥}^4+6{𝑥}^3-2{𝑥}^2+1\\ & \begin{array}{cc}& \text{ Given, }\frac{𝐻\left(𝑥\right)}{(𝑥-1)(𝑥-2)(𝑥+1)}\\ & \begin{array}{cc}& =𝑓\left(𝑥\right)+\frac{𝑔\left(𝑥\right)}{(𝑥-1)(𝑥+1)(𝑥-2}\text{ taking LCM }\\ & 𝐻\left(𝑥\right)=𝑓\left(𝑥\right)(𝑥-1)(𝑥+1)(𝑥-2)+𝑔\left(𝑥\right)\\ & 𝐻(-1)=𝑔(-1)\\ & 𝐻\left(2\right)=𝑔\left(2\right)\\ & 𝐻\left(1\right)=𝑔\left(1\right)\\ & \text{ Now, }𝐻(-1)+2𝐻\left(2\right)-3𝐻\left(1\right)\\ & =𝑔(-1)+2𝑔(2-3𝑔(1)\end{array}\end{array}\end{array}$

Given, ${630}^{\circ }<𝜃<{810}^{\circ }$ and

$\tan 𝜃=\frac{-7}{24}$

$\therefore 𝜃$ lies IV quadrant

$\begin{array}{cc}\therefore \cos 𝜃& =\frac{24}{25}{315}^{\circ }<\frac{𝜃}{2}<{405}^{\circ }\\ 2{\cos }^2\frac{𝜃}{2}-1& =\frac{24}{25}157\frac{1}{2}<\frac{𝜃}{4}<2.02\frac{1}{2}\\ 2{\cos }^2\frac{𝜃}{2}& =\frac{49}{25}\frac{𝜃}{4}\text{ lies III quadrant }\\ \Rightarrow \cos \frac{𝜃}{2}& =\frac{7}{5\sqrt{\surd 2}}\Rightarrow 2{\cos }^2\frac{𝜃}{4}-1=\frac{7}{5\sqrt{\surd 2}}\\ \Rightarrow 2{\cos }^2\frac{𝜃}{4}& =1+\frac{7}{5\sqrt{\surd 2}}=\frac{7+5\sqrt{\surd 2}}{5\sqrt{\surd 2}}\\ \Rightarrow \cos \frac{𝜃}{4}& =-\sqrt{\surd \frac{7+5\sqrt{\surd 2}}{10\sqrt{\surd 2}}}\end{array}$

We have,

$2\cos 𝜃+\sin 𝜃=1𝜃\in [-\frac{𝜋}{2},\frac{𝜋}{2}]$

Squaring both sides, we get

$\begin{array}{cc}& \Rightarrow (2\cos 𝜃{)}^2=(1-\sin 𝜃{)}^2\\ & \Rightarrow 4{\cos }^2𝜃=1+{\sin }^2𝜃-2\sin 𝜃\\ & \Rightarrow 5{\sin }^2𝜃-2\sin 𝜃-3=0\\ & \Rightarrow (\sin 𝜃-1)(5\sin 𝜃+3)=0\\ & \Rightarrow \sin 𝜃=1\text{ and }-\frac{3}{5}\\ & \Rightarrow \cos 𝜃=0\text{ and }+\frac{4}{5}\\ & \because 7\cos 𝜃+6\sin 𝜃=𝑘,\text{ when }\sin 𝜃=1\\ & \Rightarrow 7\left(0\right)+6\left(1\right)=𝑘\Rightarrow 𝑘=6\end{array}$

$\begin{array}{cc}& \text{ When, }\sin 𝜃=-\frac{3}{5}\text{ and }\cos 𝜃=\frac{4}{5}\\ & 𝑘=7\left(\frac{4}{5}\right)-6\left(\frac{3}{5}\right)=\frac{28-18}{5}=\frac{10}{5}=2\end{array}$

$\begin{array}{cc}& \sum _{𝑘=0}^{12}\frac{1}{\sin \left(\right(𝑘+1)\frac{𝜋}{6}+\frac{𝜋}{4})\cdot \sin (\frac{𝑘𝜋}{6}+\frac{𝜋}{4})}\\ & \text{ Multiplying }{𝑁}^{𝑟}\text{ and }{𝐷}^{𝑟}\text{ by }\sin 𝜋/6\text{, }\\ & \begin{array}{cc}& \frac{1}{\sin \frac{𝜋}{6}}\sum _{𝑘=0}^{12}\frac{\sin \{\left(\right(𝑘+1)\frac{𝜋}{6}+\frac{4}{𝜋})-(\frac{𝑘𝜋}{6}+\frac{𝜋}{4})\}}{2\sum _{12}^{𝑘=0}\left(\right(𝑘+1)\frac{𝜋}{6}+\frac{𝜋}{4})\cdot \sin (\frac{𝑘𝜋}{6}+\frac{𝜋}{4})}\\ & =2[\cot (\frac{𝑘𝜋}{6}+\frac{𝜋}{4})-\cot \left(\right(𝑘+1)\frac{𝜋}{6}+\frac{𝜋}{4})\}\\ & +\{\cot (\frac{𝜋}{6}+\frac{𝜋}{4})-\cot (\frac{𝜋}{6}+\frac{𝜋}{4})\}\\ & +\dots +\{\cot (\frac{12𝜋}{6}+\frac{𝜋}{4})-\cot (\frac{13𝜋}{6}+\frac{𝜋}{4})\}\\ & =2[\cot \frac{𝜋}{4}+\cot (\frac{𝜋}{6}+\frac{𝜋}{4})]\\ & =2[1-\frac{\sqrt{\surd 3}-1}{\sqrt{\surd 3}+1}]=2\left(\frac{2}{\sqrt{\surd 3}+1}\right)=\frac{4(\sqrt{\surd 3}-1)}{2}\\ & =2(\sqrt{\surd 3}-1)\end{array}\end{array}$

We have,

$2{\sin }^2𝜃-3{\cos }^2𝜃=\sin 𝜃\cos 𝜃$

Dividing both sides by ${\cos }^2𝜃$, we get $\cos 𝜃\ne 0$

$\begin{array}{cc}& \Rightarrow 2{\tan }^2𝜃-3=\tan 𝜃\\ & \Rightarrow 2{\tan }^2𝜃-\tan 𝜃-3=0\\ & \Rightarrow (2\tan 𝜃-3)(\tan 𝜃+1)=0\\ & \because \tan 𝜃=\frac{3}{2}\text{ and }\tan 𝜃=-1\end{array}$

Clearly, in the interval ( $-𝜋,𝜋$ ) $\tan 𝜃=\frac{3}{2}$ gives two solution and $\tan 𝜃=-1$ gives two solution $\therefore$ Total number of solution $=4$