Question

Find the range of values of t for which \[\]\[2\sin t = \dfrac{{1 - 2x + 5{x^2}}}{{3{x^2} - 2x - 1}}\], \[t \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]\]

Answer 1

Hint: For this type of questions first take the given quadratic expression is equal to some constant say y that is taking sint as y and make it as a single equation then find the discriminant value and solve .

Complete step-by-step answer:
The objective of the problem is to solve the range of values of t for which \[2\sin t = \dfrac{{1 - 2x + 5{x^2}}}{{3{x^2} - 2x - 1}}\] where \[t \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]\]
First ,taking the right hand side quadratic expression as some constant term say it y
That is , \[y = \dfrac{{1 - 2x + 5{x^2}}}{{3{x^2} - 2x - 1}}.....\left( 1 \right)\]
For this quadratic expression the denominator should not be zero .
Now taking the denominator to the right side in equation (1)
$\Rightarrow$ \[y\left( {3{x^2} - 2x - 1} \right) = 1 - 2x + 5{x^2}\]\[\]
$\Rightarrow$ \[3{x^2}y - 2xy - y = 1 - 2x + 5{x^2}\]
Sending the terms to the left hand side of above equation ,we get
$\Rightarrow$ \[3{x^2}y - 2xy - y - 1 + 2x - 5{x^2} = 0\]
Taking common terms of \[{x^2},x\] terms ,we get
$\Rightarrow$ \[\left( {3y - 5} \right){x^2} - 2\left( {y - 1} \right)x - \left( {y + 1} \right) = 0\]
In this expression x is real. Since x is real it should have real roots, that is its discriminate value should be greater than or equal to zero.
 That is \[D \geqslant 0\] ,\[D = {b^2} - 4ac \geqslant 0\]
Where \[a = 3y - 5,b = - 2\left( {y - 1} \right),c = - \left( {y + 1} \right)\]
Now, \[{\left( { - 2\left( {y - 1} \right)} \right)^2} - 4\left( {3y - 5} \right)\left( { - \left( {y + 1} \right)} \right) \geqslant 0\]
On computing the above equation we get
$\Rightarrow$ \[4{\left( {y - 1} \right)^2} + 4\left( {3y - 5} \right)\left( {y + 1} \right) \geqslant 0\]
On expanding the terms we get
$\Rightarrow$ \[4\left( {{y^2} - 2y + 1} \right) + 4\left( {3{y^2} + 3y - 5y - 5} \right) \geqslant 0\]
Taking four as common we get
$\Rightarrow$ \[\left( {{y^2} - 2y + 1} \right) + \left( {3{y^2} - 2y - 5} \right) \geqslant 0\]
Solving the above equation we get
\[\
  {y^2} - 2y + 1 + 3{y^2} - 2y - 5 \geqslant 0 \\
\Rightarrow 4{y^2} - 4y - 4 \geqslant 0 \\
\Rightarrow {y^2} - y - 1 \geqslant 0.....\left( 2 \right) \\
\ \]
Now we have to solve polynomial inequality. To find roots of polynomial inequality (2) we use the formula \[y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . Here \[a = 1,b = - 1,c = - 1\]
Substituting the values in above formula we get
\[y = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}\]
\[
\Rightarrow y = \dfrac{{1 \pm \sqrt {1 + 4} }}{2} \\
\Rightarrow y = \dfrac{{1 \pm \sqrt 5 }}{2} \\
 \]
Now the two roots are \[y = \dfrac{{1 + \sqrt 5 }}{2},\dfrac{{1 - \sqrt 5 }}{2}\]
Here \[y \leqslant \dfrac{{1 - \sqrt 5 }}{2}\,\,\,\,\,,\,\,y \geqslant \dfrac{{1 + \sqrt 5 }}{2}\]
Now the range of y is \[2\sin t \leqslant \dfrac{{1 - \sqrt 5 }}{2}\,\,\,\,\,2\sin t \geqslant \dfrac{{1 + \sqrt 5 }}{2}\] [since \[y = 2\sin t\]]
We can write \[ - 1 \leqslant \sin t \leqslant \dfrac{{1 - \sqrt 5 }}{4}\] or \[\dfrac{{1 + \sqrt 5 }}{4} \leqslant \sin t \leqslant 1\]
We know that \[\sin \left( {\dfrac{\pi }{2}} \right) = 1\] and also the value of \[\sin \left( {\dfrac{\pi }{{10}}} \right) = \dfrac{{\sqrt 5 - 1}}{4}\]
Now let us take \[ - 1 \leqslant \sin t \leqslant \dfrac{{1 - \sqrt 5 }}{4}\] and multiplying with hyperbolic function we get
$\Rightarrow$ \[{\sin ^{ - 1}}\left( { - 1} \right) \leqslant {\sin ^{ - 1}}\left( {\sin t} \right) \leqslant {\sin ^{ - 1}}\left( {\dfrac{{1 - \sqrt 5 }}{4}} \right)\]
\[ \Rightarrow \dfrac{{ - \pi }}{2} \leqslant t \leqslant \dfrac{{ - \pi }}{{10}}\]
Again we know that \[\cos \left( {\dfrac{{2\pi }}{{10}}} \right) = \dfrac{{1 + \sqrt 5 }}{4}\] but we need function in cos for this we need the complementary function . The complementary function is
$\Rightarrow$ \[\sin \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{{10}}} \right) = \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{5}} \right) \Rightarrow \sin \left( {\dfrac{{3\pi }}{{10}}} \right)\]
Let us consider \[\dfrac{{1 + \sqrt 5 }}{4} \leqslant \sin t \leqslant 1\] and multiplying with hyperbolic function we get
\[{\sin ^{ - 1}}\left( {\dfrac{{1 + \sqrt 5 }}{4}} \right) \leqslant {\sin ^{ - 1}}\left( {\sin t} \right) \leqslant {\sin ^{ - 1}}\left( 1 \right)\]
\[ \Rightarrow \dfrac{{3\pi }}{{10}} \leqslant t \leqslant \dfrac{\pi }{2}\]
The value of t lies between \[t \in \left[ { - \dfrac{\pi }{2}, - \dfrac{\pi }{{10}}} \right] \cup \left[ {\dfrac{{3\pi }}{{10}},\dfrac{\pi }{2}} \right]\]
We know that the fundamental period for sinx is \[2\pi \]
The generalized solution is \[t \in \left[ {2n\pi - \dfrac{\pi }{2},2n\pi - \dfrac{\pi }{{10}}} \right] \cup \left[ {2n\pi + \dfrac{{3\pi }}{{10}},2n\pi + \dfrac{\pi }{{12}}} \right]\] where \[n \in Z\]

Note: The fundamental period for sinx is \[2\pi \] and the fundamental period for cosx is also \[2\pi \] and the fundamental period for tanx is \[\pi \].If the roots of the equation is real the discriminant value should be greater than or equal to zero and if roots are equal the value should be equal to zero if the roots are imaginary the value should be less than or equal to zero.