No Problemmo #3

${ cos }^{ 2 }\left[ \frac { \pi }{ 10 } \left( 2{ cos }^{ 2 }x-5cosx+2 \right) \right] ={ sec }^{ 2 }\left( x+t{ sec }^{ 2 }x \right)$ . $\quad \quad \quad\quad \quad \quad$ Clearly LHS $\le$ 1 and RHS $\ge$ 1. $\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad$ So, LHS = RHS = 1 $\therefore { cos }^{ 2 }\left[ \frac { \pi }{ 10 } \left( 2{ cos }^{ 2 }x-5cosx+2 \right) \right] =1\quad or\quad { sin }^{ 2 }\left[ \frac { \pi }{ 10 } \left( 2{ cos }^{ 2 }x-5cosx+2 \right) \right] =0\quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \therefore \frac { \pi }{ 10 } \left( 2{ cos }^{ 2 }x-5cosx+2 \right) =n\pi \quad For\quad some\quad integer\quad n.\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\$ .

$(10n\quad =\quad 2{ cos }^{ 2 }x-5cosx+2)\\ \quad \quad \quad \quad =2\{ { cos }^{ 2 }x\quad -\frac { 5 }{ 2 } cosx\quad +1\} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad =2[{ (cosx-\frac { 5 }{ 4 } ) }^{ 2 }+1-\frac { 25 }{ 16 } ]\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad =2{ (cosx-\frac { 5 }{ 4 } ) }^{ 2 }-\frac { 9 }{ 8 } \quad \quad \quad \quad \quad \because \quad cosx\quad \in \quad [-1,1]\quad { (cosx-\frac { 5 }{ 4 } ) }^{ 2 }\quad \in \quad [\frac { 1 }{ 16 } ,\frac { 81 }{ 16 } ]\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \therefore \quad 2{ (cosx-\frac { 5 }{ 4 } ) }^{ 2 }-\frac { 9 }{ 8 } \quad \in \quad [-1,9],\quad \therefore \quad n\quad =\quad 0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ . $2{ cos }^{ 2 }x-5cosx+2\quad =\quad 0\quad \therefore cosx\quad =\quad \frac { 1 }{ 2 } \quad \therefore { sec }^{ 2 }x=4\quad or\quad x=2k\pi \pm \frac { \pi }{ 3 } \quad For\quad some\quad integer\quad k.\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ Again\quad RHS\quad =\quad 1\quad \therefore { sec }^{ 2 }\left\{ x+t{ sec }^{ 2 }x \right\} =1\quad \therefore \quad { sin }^{ 2 }\{ x+t{ sec }^{ 2 }x\} =0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ x+t{ sec }^{ 2 }x\quad =\quad l\pi \quad For\quad some\quad integer\quad l.\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ x+4t\quad =\quad l\pi ,\quad \therefore 2k\pi \pm \frac { \pi }{ 3 } \quad +\quad 4t\quad =\quad l\pi \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \therefore 4t\quad =\quad (l-2k)\pi \quad \mp \quad \frac { \pi }{ 3 } \quad \quad (l-2k)\quad can\quad be\quad be\quad assumed\quad to\quad be\quad another\quad integer\quad m.\quad \quad \quad \quad \quad \\ Hence,\quad t\quad =\quad \frac { \pi }{ 12 } \left( 3m\quad \mp \quad 1 \right) .$ .