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

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\sqrt{\surd \frac{\left[𝑥\right]-1}{[𝑥{]}^2-[𝑥]-6}}\\ & \begin{array}{cc}& 𝑓\left(𝑥\right)\text{ is defined when }\frac{\left[𝑥\right]-1}{[𝑥{]}^2-[𝑥]-6}\ge 0\\ & \frac{\left[𝑥\right]-1}{\left(\right[𝑥]-3)\left(\right[𝑥]+2)}\ge 0\\ & \Rightarrow -2<\left[𝑥\right]\le 1\text{ or }\left[𝑥\right]\ge 3\\ & \Rightarrow -1\le 𝑥<2\text{ or }𝑥\ge 4\text{ or }𝑥\in [-1,2)\cup [4,\infty )\end{array}\end{array}$

$𝑓\left(𝑥\right)=𝑥-(-1{)}^{𝑥}=\{\begin{array}{cc}𝑥-1,& 𝑥\in \text{ even }\\ 𝑥+1,& 𝑥\in \text{ odd }\end{array}$

$𝑓\left(𝑥\right)$ is a linear function. Hence, it is one-one and its range is integer for integral value of $𝑥$ so it is onto.

$\because$ It is both one-one and onto.

$\begin{array}{cc}& \text{ Given, }2\cdot 5+5\cdot 9+8\cdot 13+11\cdot 17+\dots +\text{ upto }𝑛\text{ terms }\\ & \begin{array}{cc}& {𝑡}_{𝑛}=(3𝑛-1)(4𝑛+1)\\ & {𝑡}_{𝑛}=12{𝑛}^2-𝑛-1\\ & \therefore {𝑆}_{𝑛}=\Sigma {𝑡}_{𝑛}=\Sigma (12{𝑛}^2-𝑛-1)=12\Sigma {𝑛}^2-\Sigma 𝑛-\Sigma 1\\ & =12\cdot \frac{𝑛(𝑛+1)(2𝑛+1)}{6}-\frac{𝑛(𝑛+1)}{2}-𝑛\\ & =𝑛[2(2{𝑛}^2+3𝑛+1)-\frac{𝑛+1}{2}-1]=\frac{𝑛}{2}[8{𝑛}^2+12𝑛+4-𝑛-1-2]\\ & =\frac{𝑛}{2}[8{𝑛}^2+11𝑛+1]\\ & =4{𝑛}^3+\frac{11}{2}{𝑛}^2+\frac{𝑛}{2}=𝑎{𝑛}^3+𝑏{𝑛}^2+𝑐𝑛+𝑑\\ & \Rightarrow 𝑎=4,𝑏=\frac{11}{2},𝑐=\frac{1}{2}\text{ and }𝑑=0\\ & \therefore 𝑎-𝑏+𝑐-𝑑=4-\frac{11}{2}+\frac{1}{2}-0=4-5=-1\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& -2\\ 2& -1& 2\\ -1& 1& -2\end{array}⎤⎦⎥\\ & \begin{array}{cc}\left|𝐴\right|& =\left|\right(2-2)-2(-4+2)-2(2-1\left)\right|=4-2=2\\ \mathrm{adj}\left(𝐴\right)& =⎡⎣⎢\begin{array}{ccc}0& 2& 2\\ 2& -4& -6\\ 1& -3& -5\end{array}⎤⎦⎥\\ {𝐴}^{-1}& =\frac{\mathrm{adj}\left(𝐴\right)}{\left|𝐴\right|}=\frac{1}{2}⎡⎣⎢\begin{array}{ccc}0& 2& 2\\ 2& -4& -6\\ 1& -3& -5\end{array}⎤⎦⎥\\ \text{ Then, }𝐴+2{𝐴}^{-1}& =⎡⎣⎢\begin{array}{ccc}1& 4& 0\\ 4& -5& -4\\ 0& -2& -7\end{array}⎤⎦⎥\end{array}\end{array}$

We have, $𝐴=⎡⎣⎢\begin{array}{ccc}𝑎& 𝑏& 𝑐\\ 𝑑& 𝑒& 𝑓\\ 𝑙& 𝑚& 𝑛\end{array}⎤⎦⎥$ and $\mathrm{adj}\left(𝐴\right)=⎡⎣⎢\begin{array}{ccc}0& 4& -6\\ 10& 8& 0\\ 2& 4& -4\end{array}⎤⎦⎥|\mathrm{adj}𝐴|=|𝐴{|}^2\Rightarrow |𝐴{|}^2=16\Rightarrow \left|𝐴\right|=4$

We know that $\mathrm{adj}(\mathrm{adj}𝐴)=|𝐴{|}^{𝑛-2}𝐴=|𝐴{|}^{3-2}𝐴=\left|𝐴\right|𝐴$

$\begin{array}{cc}& \Rightarrow 𝐴=\frac{1}{\left|𝐴\right|}\mathrm{adj}(\mathrm{adj}𝐴)\\ & \text{ Here, }\mathrm{adj}(\mathrm{adj}𝐴)=⎡⎣⎢\begin{array}{ccc}-32& -8& 48\\ 40& 12& -60\\ 24& 8& -40\end{array}⎤⎦⎥\\ & \therefore 𝐴=\frac{1}{4}⎡⎣⎢\begin{array}{ccc}-32& -8& 48\\ 40& 12& -60\\ 24& 8& -40\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}-8& -2& 12\\ 10& 3& -15\\ 6& 2& -10\end{array}⎤⎦⎥\\ & \begin{array}{c}\Rightarrow 𝑎=-8,𝑏=-2,𝑐=12,𝑑=10,𝑒=3,𝑓=-15,𝑙=6,𝑚=2,\\ 𝑛=-10\\ \therefore \frac{𝑐𝑑}{𝑓𝑏}+\frac{\ln }{𝑒𝑚}=\frac{12\times 10}{(-15)\times (-2)}+\frac{6\times (-10)}{3\times 2}\\ =\frac{120}{30}-\frac{60}{6}=4-10=-6\end{array}\end{array}$

Also, (a) $2𝑎=-16$

(b) $𝑎+𝑚=-8+2=-6$

(c) $𝑎+𝑏=-8-2=-10$

(d) $𝑎=-8$

$\begin{array}{cc}𝐺𝑖𝑣𝑒𝑛,{\Delta }_1& =\mid \begin{array}{ccc}-11& 1& -7\\ -4& 1& -2\\ 5& 1& 1\end{array}\mid \\ \because \Delta {\Delta }_1& =-11\left(3\right)-1\left(6\right)-7(-9)=-33-6+63=24\\ {\Delta }_3& =\mid \begin{array}{ccc}4& 1& -11\\ 1& 1& -4\\ 4& 1& 5\end{array}\mid =4\left(9\right)-1\left(21\right)-11(-3)=48\\ \Delta & =\mid \begin{array}{ccc}4& 1& -7\\ 1& 1& -2\\ 4& 1& 1\end{array}\mid =4\left(3\right)-1\left(9\right)-7(-3)=24\end{array}$

$\begin{array}{cc}& \text{ By inspection }\\ & \begin{array}{cc}& {\Delta }_2=\mid \begin{array}{ccc}4& -11& -7\\ 1& -4& -2\\ 4& 5& 1\end{array}\mid =4\left(6\right)+11\left(9\right)-7\left(21\right)=-24\\ & 𝑥=\frac{{\Delta }_1}{\Delta }=\frac{24}{24}=1,𝑦=\frac{{\Delta }_2}{\Delta }=\frac{-24}{24}=-1\text{ and }𝑧=\frac{{\Delta }_3}{\Delta }=\frac{48}{24}=2\\ & 𝑋=⎡⎣⎢\begin{array}{c}𝑥\\ 𝑦\\ 𝑧\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}1\\ -1\\ 2\end{array}⎤⎦⎥\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\left|𝑧\right|=1\\ & \begin{array}{cc}& \begin{array}{cc}& \because \text{ Let }𝑧=\cos 𝜃+𝑖\sin 𝜃={𝑒}^{𝑖𝜃}\\ & \begin{array}{cc}1+{𝑧}^2& =1+\cos 2𝜃+𝑖\sin 2𝜃\\ & =2{\cos }^2𝜃+2𝑖\sin 𝜃\cos 𝜃\\ & =2\cos 𝜃(\cos 𝜃+𝑖\sin 𝜃)=2\cos 𝜃\cdot {𝑒}^{𝑖𝜃}\end{array}\\ & \text{ Now, }\frac{1+{𝑧}^2}{2𝑖𝑧}=\frac{2\cos 𝜃\cdot {𝑒}^{𝑖𝜃}}{2𝑖{𝑒}^{𝑖𝜃}}=\frac{\cos 𝜃}{𝑖}=-𝑖\cos 𝜃\end{array}\\ & \text{ Now, Im }\left(\frac{1+{𝑧}^2}{2𝑖𝑧}\right)=-\cos 𝜃=-\mathrm{Re}\left(𝑧\right)\end{array}\end{array}$

Given, $(3+4𝑖{)}^{2025}=5^{2023}(𝑥+𝑖𝑦)$

Taking modulus on both sides, we get

$\begin{array}{cc}{5}^{2025}& =5^{2023}\sqrt{\surd {𝑥}^2+{𝑦}^2}\\ \because \sqrt{\surd {𝑥}^2+{𝑦}^2}& =5^2=25\end{array}$

$\begin{array}{cc}& \text{ We have, }{\left(\frac{\cos 𝜃+𝑖\sin 𝜃}{\sin 𝜃+𝑖\cos 𝜃}\right)}^{2024}+{\left(\frac{1+\cos 𝜃+𝑖\sin 𝜃}{1-\cos 𝜃+𝑖\sin 𝜃}\right)}^{2025}\\ & \begin{array}{cc}& ={\left\{\frac{(\cos 𝜃+𝑖\sin 𝜃)}{𝑖(\cos 𝜃-𝑖\sin 𝜃)}\right\}}^{2024}+{\left(\frac{2{\cos }^2\frac{𝜃}{2}+2𝑖\sin \frac{𝜃}{2}\cos \frac{𝜃}{2}}{2{\sin }^2\frac{𝜃}{2}+2𝑖\sin \frac{𝜃}{2}\cos \frac{𝜃}{2}}\right)}^{2025}\\ & ={\left(\frac{{𝑒}^{𝑖𝜃}}{{𝑒}^{-𝑖𝜃}}\right)}^{2024}+\frac{{(2\cos \frac{𝜃}{2})}^{2025}{(\cos \frac{𝜃}{2}+𝑖\sin \frac{𝜃}{2})}^{2025}}{{(2\sin \frac{𝜃}{2})}^{2025}{(\sin \frac{𝜃}{2}+𝑖\cos \frac{𝜃}{2})}^{2025}}\\ & ={\left({𝑒}^{2𝑖𝜃}\right)}^{2024}+{(\cot \frac{𝜃}{2})}^{2025}\frac{{\left({𝑒}^{𝑖\frac{𝜃}{2}}\right)}^{2025}}{{𝑖}^{2025}{\left({𝑒}^{-𝑖\frac{𝜃}{2}}\right)}^{2025}}\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& ={\left({𝑒}^{2𝑖𝜃}\right)}^{2024}+{(\cot \frac{𝜃}{2})}^{2025}\frac{1}{𝑖}{\left({𝑒}^{𝑖𝜃}\right)}^{2025}\\ & =(\cos 2𝜃+𝑖\sin 2𝜃{)}^{2024}-{(\cot \frac{𝜃}{2})}^{2025}𝑖(\cos 𝜃+𝑖\sin 𝜃{)}^{2025}\end{array}\\ & \text{ Now, put }𝜃=\frac{𝜋}{2}\\ & =(-1{)}^{2024}-𝑖(𝑖{)}^{2025}=1-{𝑖}^2=1+1=2\end{array}$

Given, ${𝑥}^2-6(𝑘-1)𝑥+4(𝑘-2)=0$

$\begin{array}{cc}& 𝛼+𝛽=6(𝑘-1)=0\Rightarrow 𝑘=1\\ & 𝛼𝛽=4(𝑘-2)<0\Rightarrow 𝑘<2\end{array}$

for $𝑘=1,{𝑥}^2-4=0\Rightarrow {𝑥}^2=4$

$\therefore 𝑥=\pm 2\Rightarrow 𝛼=+2\text{ and }𝛽=-2$

Then, $2{𝑥}^2-𝛼𝑥+6𝛽(𝛼+1)=0$ become

$\begin{array}{c}2{𝑥}^2-2𝑥+6(-2)\left(3\right)=0\\ {𝑥}^2-𝑥-18=0\end{array}$

Now, product of roots $=-18$

We have, $𝑎{𝑥}^2+𝑏𝑥+𝑐<0$

If $𝑎<0$, then $\Delta <0$

$𝑎<0\text{ and }{𝑏}^2-4𝑎𝑐<0$

$\therefore {𝑏}^2<4𝑎𝑐$

And $𝑐{𝑥}^2+𝑎𝑥+𝑏$ and $𝑎{𝑥}^2+𝑏𝑥+𝑐$ have same extreme value

$\because -\frac{𝑎}{2𝑐}=-\frac{𝑏}{2𝑎}\Rightarrow {𝑎}^2=𝑏𝑐$

Now, the minimum value of $𝑐{𝑥}^2+𝑎𝑥+𝑏$ is $-\frac{\Delta }{4𝑎}=-\frac{{𝑎}^2-4𝑏𝑐}{4𝑐}=-\frac{𝑏𝑐-4𝑏𝑐}{4𝑐}=-(-\frac{3𝑏𝑐}{4𝑐})=\frac{3𝑏}{4}$

Because ${𝑏}^2=$ positive and $𝑎<0\Rightarrow 𝑐<0$

$\therefore \frac{3𝑏}{4}$ is the maximum value of $𝑐{𝑥}^2+𝑎𝑥+𝑏$

Given,

${𝑥}^4-10{𝑥}^3+50{𝑥}^2-130𝑥+169=0$

Sum of roots $2𝑎+2𝑏=10\Rightarrow 𝑎+𝑏=5...\left(𝑖\right)$

Product of roots $({𝑎}^2+{𝑏}^2)({𝑏}^2+{𝑎}^2)=169$

${𝑎}^2+{𝑏}^2=13...\left(𝑖𝑖\right)$

From, Eq.s (i) and (ii) $𝑎𝑏=6$

$\frac{{𝑎}^2+{𝑏}^2}{𝑎𝑏}=\frac{13}{6}\Rightarrow \frac{𝑎}{𝑏}+\frac{𝑏}{𝑎}=\frac{13}{6}$

$𝑓\left(𝑥\right)={𝑥}^4-17{𝑥}^3+𝑘{𝑥}^2-247𝑥+210$

Its factor: ${𝑥}^2-5𝑥+6$

$\because (𝑥-2)(𝑥-3)$ is also a factor.

$\begin{array}{cc}& f\left(2\right)=16-17\left(8\right)+𝑘\left(4\right)-\left(247\right)2+210=0\\ & \Rightarrow 16-136+4𝑘-494+210=0\\ & \Rightarrow 4𝑘-404=0\Rightarrow 𝑘=101\\ & \because 𝑓\left(𝑥\right)={𝑥}^4-17{𝑥}^3+101{𝑥}^2-247𝑥+210\\ & =({𝑥}^2-5𝑥+6)({𝑥}^2-12𝑥+35)\end{array}$

COMBINATION

Vowels- O, O, I, I, A = 5

$\begin{array}{cc}& =4\times \frac{8!}{2!2!}\times 2\left(C\text{ and }N\text{ are also be arranged }\right)\\ & =2\times 8!\end{array}$

Total number of fruits $=3\times 12=36$

These fruits distributed among 9 persons equals $\because$ Order is taken'.

Total number of ways $=\frac{36!}{(4!{)}^{9}9!}\times 9!=\frac{36!}{(4!{)}^9}$.

We have, $\sum _{𝑖=0}^{𝑚}{}^{10}{𝐶}_{𝑖}{}^{20}{𝐶}_{𝑚-𝑖}$

$={}^{10}{𝐶}_0{}^{20}{𝐶}_{𝑚}+{}^{10}{𝐶}_1{}^{20}{𝐶}_{𝑚-1}+{}^{10}{𝐶}_2{}^{20}{𝐶}_{𝑚-2}+\dots +{}^{10}{𝐶}_{𝑚}\cdot {}^{20}{𝐶}_0$

Coefficient of ${𝑥}^{𝑚}$ in the expansion of the product

$(1+𝑥{)}^{10}(1+𝑥{)}^{20}$

$=$ Coefficient of ${𝑥}^{𝑚}$ in $(1+𝑥{)}^{30}$

$={}^{30}{𝐶}_{𝑚}.$

Maximum value of ${}^{30}{𝐶}_{𝑚}={}^{30}{𝐶}_{15}$

$\therefore 𝑚=15$

${({𝑥}^2+𝑥-2)}^5$

Apply multinomial theorem

$\Sigma \frac{5!}{𝑝!𝑞!𝑟!}{\left({𝑥}^2\right)}^{𝑝}\left(𝑥{)}^{𝑞}\right(-2{)}^{𝑟}$

Such that $𝑝+𝑞+𝑟=5$ and $𝑝,𝑞,𝑟$ are non-negative integer.

$\begin{array}{cc}& 2𝑝+𝑞=2\\ & (𝑝,𝑞,𝑟)=(1,0,4)(0,2,3)\\ & \frac{5!}{4!}(-4^4+\frac{5!}{2!3!}(-2^3\\ & 5\times 16+10(-8)=80-80=0\end{array}$

We have, ${𝑃}_{𝑛}={}^{𝑛}{𝐶}_0\cdot {}^{𝑛}{𝐶}_1{}^{𝑛}{𝐶}_2\dots {}^{𝑛}{𝐶}_{𝑛}$ and

$\begin{array}{cc}& {𝑃}_{𝑛+1}={}^{𝑛+1}{𝐶}_0\cdot {}^{𝑛+1}{𝐶}_1{}^{𝑛+1}{𝐶}_2\cdots {}^{𝑛+1}{𝐶}_{𝑛+1}\\ & \therefore \frac{{𝑃}_{𝑛+1}}{{𝑃}_{𝑛}}=\frac{{}^{𝑛+1}{𝐶}_0\cdot {}^{𝑛+1}{𝐶}_1{}^{𝑛+1}{𝐶}_2\cdots {}^{𝑛+1}{𝐶}_{𝑛+1}}{{}^{𝑛}{𝐶}_0\cdot {}^{𝑛}{𝐶}_1\cdot {}^{𝑛}{𝐶}_2\cdots {}^{𝑛}{𝐶}_{𝑛}}\end{array}$

$\begin{array}{cc}& =\left(\frac{{}^{𝑛+1}{𝐶}_1}{{}^{𝑛}{𝐶}_0}\right)\left(\frac{{}^{𝑛+1}{𝐶}_2}{{}^{𝑛}{𝐶}_1}\right)\left(\frac{{}^{𝑛+1}{𝐶}_3}{{}^{𝑛}{𝐶}_2}\right)\cdots \left(\frac{{}^{𝑛+1}{𝐶}_{𝑛+1}}{{}^{𝑛}{𝐶}_{𝑛}}\right)\\ & =\left(\frac{𝑛+1}{1}\right)\left(\frac{𝑛+1}{2}\right)\left(\frac{𝑛+1}{3}\right)\cdots \left(\frac{𝑛+1}{𝑛+1}\right)\\ & =\frac{(𝑛+1{)}^{𝑛+1}}{(𝑛+1)!}\end{array}$

Given, $\frac{{𝑥}^4+1}{({𝑥}^2+1)(𝑥-1)}$

$\begin{array}{cc}& =\frac{{𝑥}^4(1+\frac{1}{{𝑥}^4})}{{𝑥}^3(1+\frac{1}{{𝑥}^2})(1-\frac{1}{𝑥})}\\ & =𝑥(1+\frac{1}{{𝑥}^4}){(1+\frac{1}{{𝑥}^2})}^{-1}{(1-\frac{1}{𝑥})}^{-1}\\ & =𝑥(1+\frac{1}{{𝑥}^4})(1-\frac{1}{{𝑥}^2})(1+\frac{1}{𝑥}+\frac{1}{{𝑥}^2}+\dots )\end{array}$

Clearly, there is no terms containing ${𝑥}^3$ coefficient of ${𝑥}^3=0$

$\begin{array}{cc}& \text{ We have, }\sum _{𝑟=1}^{89}\frac{1}{\sin {𝑟}^{\circ }\cdot \sin (𝑟+1{)}^{\circ }}\\ & \begin{array}{cc}& \Rightarrow \frac{1}{\sin 1^{\circ }}\sum _{𝑟=1}^{89}\frac{\sin (({𝑟}^{\circ }+1^{\circ })-{𝑟}^{\circ })}{\sin {𝑟}^{\circ }\cdot \sin ({𝑟}^{\circ }+1^{\circ })}\\ & \Rightarrow \frac{1}{\sin 1^{\circ }}\sum _{𝑟=1}^{89}\frac{\sin (𝑟+1{)}^{\circ }\cos {𝑟}^{\circ }}{\sin {𝑟}^{\circ }\cdot \sin (𝑟+1{)}^{\circ }}\\ & \Rightarrow \frac{1}{\sin 1^{\circ }}\sum _{𝑟=1^{\circ }}^{89}\cot {𝑟}^{\circ }-\cot (𝑟+1{)}^{\circ }\\ & \Rightarrow \frac{1}{\sin 1^{\circ }}[\cot 1^{\circ }-\cot {90}^{\circ }]=\frac{\cot 1^{\circ }}{\sin 1^{\circ }}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}& {\cos }^3\frac{𝜋}{8}\cdot \cos \frac{3𝜋}{8}+{\sin }^3\frac{𝜋}{8}\cdot \sin \frac{3𝜋}{8}\\ & \text{ Using }\frac{3𝜋}{8}=\frac{𝜋}{2}-\frac{𝜋}{8}\\ & \Rightarrow \cos \frac{3𝜋}{8}=\sin \frac{𝜋}{8}\text{ and }\sin \frac{3𝜋}{8}=\cos \frac{𝜋}{8}\\ & \Rightarrow {\cos }^3\frac{𝜋}{8}\cdot \sin \frac{𝜋}{8}+{\sin }^3\frac{𝜋}{8}\cdot \cos \frac{𝜋}{8}\\ & =\sin \frac{𝜋}{8}\cdot \cos \frac{𝜋}{8}({\cos }^2\frac{𝜋}{8}+{\sin }^2\frac{𝜋}{8})\\ & =\frac{1}{2}2\sin \frac{𝜋}{8}\cos \frac{𝜋}{8}\\ & =\frac{1}{2}\sin \frac{𝜋}{4}=\frac{1}{2\sqrt{\surd 2}}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }𝐴+𝐵+𝐶=\frac{𝜋}{4}\\ & \begin{array}{cc}& \therefore 2𝐴+2𝐵+2𝐶=\frac{𝜋}{2}\\ & \text{ Now, }\sin 4𝐴+\sin 4𝐵+\sin 4𝐶\\ & 2\sin (2𝐴+2𝐵)\cdot \cos (2𝐴-2𝐵)+2\sin 2𝐶\\ & =2\sin (\frac{𝜋}{2}-2𝐶)\cdot \cos 2𝐶+2\sin 2𝐶\cdot \cos 2𝐶\\ & =2\cos 2𝐶\cdot \cos (2𝐴-2𝐵)+2\sin 2𝐶\cdot \cos 2𝐶\\ & =2\cos 2𝐶\{\cos (2𝐴-2𝐵)+\sin 2𝐶\}\\ & =2\cos 2𝐶[\cos (2𝐴-2𝐵)+\cos (2𝐴+2𝐵\left)\right]\\ & =2\cos 2𝐶[2\cos 2𝐴\cdot \cos 2𝐵]\\ & =4\cos 2𝐴\cdot \cos 2𝐵\cdot \cos 2𝐶\end{array}\end{array}$

We have,

$\begin{array}{cc}& \begin{array}{c}\cos 𝜃+\cos 2𝜃-\sqrt{\surd 3}(\sin 𝜃+\sin 2𝜃)+1=0\\ \Rightarrow \cos 𝜃+2{\cos }^2𝜃-\sqrt{\surd 3}\sin 𝜃-2\sqrt{\surd 3}\sin 𝜃\cos 𝜃=0\\ \Rightarrow (\cos 𝜃-\sqrt{\surd 3}\sin 𝜃)+(2{\cos }^2𝜃-2\sqrt{\surd 3}\sin 𝜃\cos 𝜃)=0\end{array}\\ & \begin{array}{c}\Rightarrow 1(\cos 𝜃-\sqrt{\surd 3}\sin 𝜃)+2\cos 𝜃(\cos 𝜃-\sqrt{\surd 3}\sin 𝜃)=0\end{array}\\ & \Rightarrow (\cos 𝜃-\sqrt{\surd 3}\sin 𝜃)(2\cos 𝜃+1)=0\\ & \Rightarrow \cos 𝜃-\sqrt{\surd 3}\sin 𝜃=0\text{ or }2\cos 𝜃=-1\\ & \begin{array}{c}\tan 𝜃=\frac{1}{\sqrt{\surd 3}}\text{ or }\cos 𝜃=-\frac{1}{2}=\cos \frac{2𝜋}{3}\\ 𝜃=𝑛𝜋+\frac{𝜋}{6}\text{ or }𝜃=2𝑛𝜋+\frac{2𝜋}{3}\\ 𝜃=\frac{𝜋}{6},\frac{7𝜋}{6}\text{ or }𝜃=\frac{2𝜋}{3},\frac{4𝜋}{3}\end{array}\end{array}$