Maths M.C.Q (1 Marks)

$ \Rightarrow $$2\sin 2x\cos x - 3\sin 2x - 2\cos 2x\cos x + 3\cos 2x = 0$

$ \Rightarrow $ $\sin 2x(2\cos x - 3) - \cos 2x(2\cos x - 3) = 0$

$ \Rightarrow $$(\sin 2x - \cos 2x)\,\,(2\cos x - 3) = 0 \Rightarrow \sin 2x = \cos 2x$

                                                                   $(\,\because \cos x \ne 3/2)$

$ \Rightarrow $ $2x = 2n\pi \pm \left( {\frac{\pi }{2} - 2x} \right)$

$i.e.,$ $x = \frac{{n\pi }}{2} + \frac{\pi }{8}$.

$\Rightarrow \frac{1}{{\sqrt 2 }}\sin \theta + \frac{1}{{\sqrt 2 }}\cos \theta = \frac{1}{{\sqrt 2 }}$

Dividing by $\sqrt {{1^2} + {1^2}} = \sqrt 2 $, we get

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

$ \Rightarrow $ $\theta + \frac{\pi }{4} = n\pi + {( - 1)^n}\frac{\pi }{4}$

$\Rightarrow \theta = n\pi + {( - 1)^n}\frac{\pi }{4} - \frac{\pi }{4}$.

$ \Rightarrow $ $5x = 2n\pi \pm (\pi - x)$

$ \Rightarrow $ $x = \frac{{(2n + 1)\pi }}{6}{\rm{ or }}\frac{{(2n - 1)\pi }}{4}$

Hence $x = \frac{\pi }{4},\,\frac{\pi }{2},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{6},\,\frac{{5\pi }}{4},\,\frac{{7\pi }}{6},\,\frac{{7\pi }}{4},\,\frac{{9\pi }}{6},\,\frac{{11\pi }}{6}$.

$ \Rightarrow $ $\cos 2\theta - \cos 4\theta = 2\cos 4\theta \cos 2\theta $

$ \Rightarrow $ $ - \cos 4\theta = \cos 6\theta $

$ \Rightarrow $ $2\cos 5\theta \cos \theta = 0$

$\theta = n\pi + {\pi\over{2}}$ or $\frac{{n\pi }}{5} + \frac{\pi }{{10}}$.

==> $2{\cos ^2} + 3\cos \theta - 2 = 0$

==> $\cos \theta = \frac{{ - 3 \pm \sqrt {9 + 16} }}{4} = \frac{{ - 3 \pm 5}}{4}$

Neglecting $(-)$ sign, we get

$\cos \theta = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right)$

$ \Rightarrow $ $\theta = 2n\pi \pm \frac{\pi }{3}$.

The values of $\theta $ between $0$ and $2\pi $ are $\frac{\pi }{3},{\rm{ }}\frac{{{\rm{5}}\pi }}{{\rm{3}}}$.

$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.

$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.

$\Rightarrow 2{\tan ^2}\theta = {\tan ^2}\theta + 1$

$ \Rightarrow $ ${\tan ^2}\theta = 1 = {\tan ^2}\left( {\frac{\pi }{4}} \right)$

$\Rightarrow \theta = n\pi \pm \frac{\pi }{4}$.

$\Rightarrow \sin \theta + \cos \theta = 1$

$ \Rightarrow $ $\cos \left( {\theta - \frac{\pi }{4}} \right)\, = \cos \frac{\pi }{4} $

$\Rightarrow \theta - \frac{\pi }{4} = 2n\pi \pm \frac{\pi }{4}$

Hence $\theta = 2n\pi $ or $\theta = 2n\pi + \frac{\pi }{2}$.

But $\theta = 2n\pi $ is ruled out.

==> $\frac{{1 - (1 - 2{{\sin }^2}\theta )}}{{1 + (2{{\cos }^2}\theta - 1)}} = 3$

$ \Rightarrow $ ${\tan ^2}\theta = 3$

$ \Rightarrow $ $\theta = n\pi \pm \frac{\pi }{3}$.

$ \Rightarrow $ $\cos \theta = \frac{{(5/2) \pm \sqrt {(25/4) - 4} }}{2} = \frac{{5 \pm 3}}{4}$

Rejecting $(+)$ sign,

$ \Rightarrow $ $\cos \theta = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right) $

$\Rightarrow \theta = 2n\pi \pm \frac{\pi }{3}$.

$\Rightarrow \sin \theta - \sqrt 3 \cos \theta = \sqrt 2 $

$ \Rightarrow $ $\sin \left( {\theta - \frac{\pi }{3}} \right) = \sin \frac{\pi }{4}$

$\Rightarrow \theta = n\pi + {( - 1)^n}\frac{\pi }{4} + \frac{\pi }{3}$.

$\Rightarrow \cos \left( {\theta - \frac{\pi }{4}} \right) = \cos \alpha $

$ \Rightarrow $ $\theta - \frac{\pi }{4} = 2n\pi \pm \alpha $

$\Rightarrow \theta = 2n\pi + \frac{\pi }{4} \pm \alpha $.

$\Rightarrow \tan \theta = \tan \left( {\frac{\pi }{2} - \alpha } \right)$

$ \Rightarrow $ $\theta = n\pi + \frac{\pi }{2} - \alpha $.

$ \Rightarrow $ ${\sin ^2}x = \frac{1}{2} = {\sin ^2}\frac{\pi }{4}$

$\Rightarrow x = n\pi \pm \frac{\pi }{4}$.

$\Rightarrow 2\theta = 2n\pi \pm \left( {\frac{\pi }{2} - \alpha } \right)$

$ \Rightarrow $ $\theta = n\pi \pm \left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$.

$\Rightarrow (\sin \theta - 1)\,(\sin \theta + 2) = 0$

$ \Rightarrow $ $\sin \theta \ne - 2$ ,

$\therefore \,\,\sin \theta = 1 = \sin \pi /2$

$ \Rightarrow $ $\theta = n\pi + {( - 1)^n}\frac{\pi }{2}$.

$ \Rightarrow $ $5\theta = n\pi + \frac{\pi }{2} - 2\theta $

$ \Rightarrow $ $7\theta = n\pi + \frac{\pi }{2}$

$ \Rightarrow $ $\theta = \frac{{n\pi }}{7} + \frac{\pi }{{14}}$.

$ \Rightarrow $ $3\theta = 2n\pi \pm \left( {\frac{\pi }{2} - 2\theta } \right)$

$ \Rightarrow $ $\theta = \frac{{2n\pi }}{5} + \frac{\pi }{{10}}$

or $\theta = 2n\pi - \frac{\pi }{2}$.

Since $\theta $ is acute

==> $\theta = \frac{\pi }{{10}}$

==> $\sin \theta = \frac{{\sqrt 5 - 1}}{4}$.

==> $\frac{{\tan 3\theta - \tan (\pi /4)}}{{1 + \tan 3\theta \,.\,\tan (\pi /4)}} = \sqrt 3 $

==> $\tan \,\left( {3\theta - \frac{\pi }{4}} \right) = \tan \,\frac{\pi }{3}$

==> $3\theta - (\pi /4) = n\pi + (\pi /3)$

==> $3\theta = n\pi + \frac{{7\pi }}{{12}}$

==> $\theta = \frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}$.

$ \Rightarrow $$(2\sin x - 1)\,(\sin x - 1) = 0$

$ \Rightarrow $ $\sin x = \frac{1}{2}$ or $\sin x = 1$

$ \Rightarrow $ $x = \frac{\pi }{6},\,\frac{{5\pi }}{6}$, $\frac{\pi }{2}$ 

$i.e.,$ $30^\circ ,{\rm{ 150}}^\circ {\rm{, 90}}^\circ $.

Aliter : Since the maximum value of $(\sin x + \cos x) =  \sqrt {{1^2} + {1^2}} = \sqrt 2 $. 

Hence there is no satisfying $\sin x + \cos x = 2$.

$ \Rightarrow $ $2{\cos ^2}\theta + 3\cos \theta + 2 = 0$

$ \Rightarrow $ $\cos \theta = \frac{{ - 3 \pm \sqrt {9 - 16} }}{4}$,

which is imaginary, hence no solution.

Also we have ${\sec ^2}\theta - {\tan ^2}\theta = 1$...$(ii)$

$ \Rightarrow $ $\sec \theta - \tan \theta = \frac{1}{{\sqrt 3 }}$...$(iii)$

Now $(i)$ and $(iii)$ gives,

$\tan \theta = \frac{1}{2}\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) = \frac{1}{{\sqrt 3 }} = \tan \left( {\frac{\pi }{6}} \right)$

$ \Rightarrow $ $\theta = n\pi + \frac{\pi }{6}$.

$\therefore $ Solutions for $0 \le \theta \le 2\pi $ are $\frac{\pi }{6}$ and $\frac{{7\pi }}{6}$.

Hence there are two solutions.

which is of the form $a\sin x + b\cos x = c$ with $a = \sqrt 3 ,\,b = 1,\,c = 4$.

Here ${a^2} + {b^2} = 3 + 1 = 4 < {c^2},$

therefore the given equation has no solution.

==> $\frac{3}{5}\cos x + \frac{4}{5}\sin x = \frac{6}{5}$

==> $\cos (x - \theta ) = \frac{6}{5}$,

$[{\rm{where\,\, }}\theta = {\cos ^{ - 1}}(3/5)]$

So, that equation has no solution.

$x = y = z = \frac{{3\pi }}{2},$ for $x,\,y,\,z \in [0,\,\,2\pi ].$

$\Rightarrow \cos \theta = \cos \left( {\frac{\pi }{2} - 2\theta } \right)$

$ \Rightarrow $ $\theta = 2n\pi \pm \left( {\frac{\pi }{2} - 2\theta } \right) $

$\Rightarrow \theta \pm 2\theta = 2n\pi \pm \frac{\pi }{2}$

$i.e.$, $3\theta = 2n\pi + \frac{\pi }{2} $

$\Rightarrow \theta = \frac{1}{3}\left( {2n\pi + \frac{\pi }{2}} \right)$

and $ - \theta = 2n\pi - \frac{\pi }{2}$

$\Rightarrow \theta = - \left( {2n\pi - \frac{\pi }{2}} \right)$

Hence value of $\theta $ between $0$ and $\pi $ are $\frac{\pi }{6},\,\frac{\pi }{2},\,\frac{{5\pi }}{6}$

$i.e.$, ${30^o},\,{90^o},\,{150^o}$.

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

$ \Rightarrow $ $\theta = 210^\circ $ or $330^\circ $.

$=  \frac{{{{(1 + {{\tan }^2}x)}^2} - {{\tan }^2}x}}{{{{\tan }^2}x}}$

Obviously, $1 + {\tan ^2}x \ge {\tan ^2}x,{\rm{ }}\forall {\rm{ }}x$. 

Hence it is positive for all value of $x.$

$\Rightarrow \theta = n\pi + \frac{\pi }{3}$

For $ - \pi < \theta < 0$,  ${\rm{Put}}\,\,n = - 1$, we get

$\theta = - \pi + \frac{\pi }{3} = \frac{{ - 2\pi }}{3}{\rm{or }}\frac{{ - 4\pi }}{6}$.

$\because$ $\theta $ lies in between $0^\circ $ and $360^\circ $

$\therefore $ $\theta = 150^\circ $ and $330^\circ $.

$\cos \theta = \frac{1}{{\sqrt 2 }} = \cos \left( {2\pi - \frac{\pi }{4}} \right)$

Hence general value is $2n\pi + \left( {2\pi - \frac{\pi }{4}} \right) = 2n\pi + \frac{{7\pi }}{4}$.

$\tan \theta = \frac{1}{{\sqrt 3 }} = \tan \left( {\frac{\pi }{6}} \right) = \tan \left( {\pi + \frac{\pi }{6}} \right)$

$\Rightarrow \theta = \left( {\pi + \frac{\pi }{6}} \right)$

Hence general value of $\theta $ is $2n\pi + \frac{{7\pi }}{6}$.

$\because$  $\sin \theta = \frac{1}{2},\, - \frac{3}{2}$ (rejected)

$ \Rightarrow $ $\theta = \frac{\pi }{6},\,\pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$.

$\tan \theta = 1 \Rightarrow \theta = \frac{\pi }{4},\,\frac{{5\pi }}{4}$

$\therefore $ The general value is $2n\pi + \frac{{5\pi }}{4}$ or $(2n + 1)\pi + \frac{\pi }{4}$.

$ \Rightarrow $ $A + B = \frac{\pi }{2}$ and $A - B = \frac{\pi }{6}$

$ \Rightarrow $ $A = \frac{\pi }{3},\,B = \frac{\pi }{6}$.

$ \Rightarrow $  $2{\cos ^2}\theta  - \sqrt 3 \cos \theta  - 3 = 0$

$ \Rightarrow $  $\cos \theta  = \frac{{\sqrt 3  \pm \sqrt {3 + 24} }}{4} = \frac{{\sqrt 3 (1 \pm 3)}}{4} = \sqrt 3 \left( { - \frac{1}{2}} \right)$

$ \Rightarrow $  $\theta  = \frac{{5\pi }}{6}$.

$\Rightarrow 2{\sin ^2}\theta \cos \theta = \cos \theta $

$ \Rightarrow $ $\cos \theta = 0$ or ${\sin ^2}\theta = \frac{1}{2} = {\sin ^2}\left( {\frac{\pi }{4}} \right)$

$ \Rightarrow $ $\theta = (2n + 1)\frac{\pi }{2}$ or $\theta = n\pi \pm \frac{\pi }{4}$

$ \Rightarrow $ $\theta = {90^o}$ and ${45^o}$.

$\Rightarrow \theta = \frac{\pi }{2} - \frac{\pi }{6} = \frac{\pi }{3}$.

$= - (2{\cos ^2}{40^o} - 1)$ $ = - \cos (2 \times {40^o})$

$=  - \cos {80^o}$ = $\cos ({180^o} + {80^o}) = \cos ({180^o} - {80^o})$ 

Hence, $\cos 260^\circ {\rm{and}}\cos 100^\circ $ 

$i.e.$, $\theta = 100^\circ $ and $260^\circ$.

$\therefore $ $\,\sin \theta + \cos \theta = \frac{1}{2}$

$ \Rightarrow \,\cos \left( {\theta - \frac{\pi }{4}} \right) = \frac{1}{{2\sqrt 2 }}$.

$\therefore \sin \theta + \cos \theta = \frac{1}{2}$

==>$\sin \left( {\theta + \frac{\pi }{4}} \right) = \frac{1}{{2\sqrt 2 }}$.

$\Rightarrow \cos \left( {x + \frac{\pi }{4}} \right) = - 1$

==> $x + \frac{\pi }{4} = 2n\pi \pm \pi $

$\Rightarrow 2n\pi + \frac{{3\pi }}{4}{\rm{\,\,or\,\,}}\,\,2n\pi - \frac{{5\pi }}{4}$ .

$ \Rightarrow $ $2{\sin ^2}\theta = 1$

$ \Rightarrow $ ${\sin ^2}\theta = \frac{1}{2} = {\sin ^2}\left( {\frac{\pi }{4}} \right) $

$\Rightarrow \theta = n\pi \pm \frac{\pi }{4}$.

$\therefore $ $\sin 3x = 0$, $\cos x = 1$

$\Rightarrow 3x = n\pi $ or $x = \frac{{n\pi }}{3}$ and $x = 2n\pi $

The second value $x = 2n\pi $ is included in the value given by $x = \frac{{n\pi }}{3}$.

$\cos \theta = - 5/4$, which is not possible.

$\therefore 2\cos \theta + 1 = 0$ or $\cos \theta = - 1/2$

==> $\theta = \frac{{2\pi }}{3},\,\,\frac{{4\pi }}{3}.$

Solution set is $\left\{ {\frac{{2\pi }}{3},\,\,\frac{{4\pi }}{3}} \right\} \in [0,\,\,2\pi ]$.

$\sin (x + \alpha ) = \frac{c}{{\sqrt {{a^2} + {b^2}} }} > 1$, (as given)

Hence there is no solution.