Question

Solve the given cubic equation:- $16{{\cos }^{6}}x-25{{\cos }^{4}}x+11{{\cos }^{2}}x-2=0$

Answer 1

Hint: Take $\text{co}{{\text{s}}^{\text{2}}}\text{x=t}$to form a cubic equation. The reduced cubic equation’s one factor can be formed by using the hit and trial method and reducing it to a quadratic equation. Solve this quadratic equation by making a factor.
Formula used:
By substitution find one factor of cubic equation & then make factors of a quadratic equation.
$\begin{align}
  & a{{x}^{2}}+bx+c=0 \\
 & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}$

Complete step-by-step answer:
$16{{\cos }^{6}}x-25{{\cos }^{4}}x+11{{\cos }^{2}}x-2=0$
Put
$\begin{align}
  & {{\cos }^{2}}x=t \\
 & 16\left[ {{t}^{3}} \right]-25\text{ }{{t}^{2}}+11\text{ }t-2=0......(1) \\
 & \text{By substitution we find one factor of cubic equation}\text{.} \\
\end{align}$
Put
$t=1$ in equation (1)
$\begin{align}
  & 16{{\left( 1 \right)}^{3}}-25{{\left( 1 \right)}^{2}}+11\left( 1 \right)-2 \\
 & =16-25+11-2 \\
 & =27-27=0 \\
\end{align}$
Therefore t = 1 is the solution of cubic equation $16\text{ }{{t}^{3}}-25\text{ }{{t}^{2}}+11\text{ }t-2$.
 [By long division]
$\text{16 }{{t}^{3}}-25\text{ }{{t}^{2}}+11\text{ t}-2=\left( t-1 \right)\left( 16\text{ }{{t}^{2}}-9t+2 \right)$
Now we will solve the quadratic equation
$\left( 16\text{ }{{t}^{2}}-9t+2 \right)$
\[\begin{align}
  &\Rightarrow 16\text{ }{{t}^{2}}-9\text{ }t+2=0 \\
 & \Rightarrow a=16\text{ } \\
 &\Rightarrow b=-9 \\
 & c=2 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow t =\dfrac{-\left( -9 \right)\pm \sqrt{{{\left( -9 \right)}^{2}}-4\times 16\times 2}}{2\times 16} \\
 & \Rightarrow t =\dfrac{9\pm \sqrt{81-128}}{32} \\
 &\Rightarrow t=\dfrac{9\pm \sqrt{-47}}{32}\text{ No, real roots because}\sqrt{-47}\text{ is not a real number} \\
 & \text{Hence 16 }{{t}^{3}}-25\text{ }{{t}^{2}}+11\text{ }t-2=\left( t-1 \right)\left( 16\,{{t}^{2}}-9t+2 \right) \\
\end{align}\]
\[\text{Substitute }=t={{\cos }^{2}}~x\]
$\begin{align}
& \therefore \left( {{\cos }^{2}}~x-1 \right)\left( 16{{\cos }^{4}}~x-9{{\cos }^{2}}~x+2 \right)=0 \\
 & \Rightarrow {{\cos }^{2}}~x-1=0\text{ or }16{{\cos }^{4}}~x-9{{\cos }^{2}}~x+2=0 \\
 &\Rightarrow {{\cos }^{2}}~x=1 \\
 & \Rightarrow {{\cos }^{2}}~x=\pm 1 \\
\end{align}$

Additional information:
 We can solve quadratic problem by splitting the middle term if we are not able to solve the quadratic problem by splitting the middle term we can use $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ formula.

Note: For higher powers we have to substitute like in this power was 6. So we substitute ${{\cos }^{2}}~x=t$ so we get a cubic calculation then factorize.