Question

The value of x which satisfy the equation ${{\sin }^{2015}}y=\left| {{x}^{3}}+{{x}^{2}}-x-1 \right|+{{e}^{\left| {{x}^{3}}-4x-{{x}^{2}}+4 \right|}}+{{\tan }^{2}}2y+{{\cos }^{4}}y$ are

Answer 1

To obtain the value of x which satisfies the given equation we will use the interval of sine function. Firstly as we know that the sine function value on the left hand side ranges from $\left[ -1,1 \right]$ then we will get the idea of what our right hand side value will be. Accordingly we will get our desired answer.

Complete step-by-step solution:
We have to find the value of x which satisfies the equation below:
${{\sin }^{2015}}y=\left| {{x}^{3}}+{{x}^{2}}-x-1 \right|+{{e}^{\left| {{x}^{3}}-4x-{{x}^{2}}+4 \right|}}+{{\tan }^{2}}2y+{{\cos }^{4}}y$
Now as we know that sine function value is from $\left[ -1,1 \right]$ when the power of sine is odd so we can say that,
$-1\le {{\sin }^{2015}}\le 1$
Next we know that the value of a function in mod is always positive and exponent to power 0 is 1.
Then we know that the following condition should hold true for the right hand side to be equal to left hand side:
$\begin{aligned} & \sin y=1 \\\ & \tan 2y=0 \\\ & \cos y=0 \\\ & \left| {{x}^{3}}+{{x}^{2}}-x-1 \right|=0 \\\ & \left| {{x}^{3}}-4x-{{x}^{2}}+4 \right|=0 \\\ \end{aligned}$
Now as we have to find the value of $x$ we will take the condition with $x$ variable in it and simplify them to get a common value in them as follows:
Firstly we will take the below equation and solve it:
$\begin{aligned} & \left| {{x}^{3}}+{{x}^{2}}-x-1 \right|=0 \\\ & \Rightarrow {{x}^{3}}+{{x}^{2}}-x-1=0 \\\ & \Rightarrow {{x}^{2}}\left( x+1 \right)-1\left( x+1 \right)=0 \\\ & \Rightarrow \left( {{x}^{2}}-1 \right)\left( x+1 \right)=0 \\\ \end{aligned}$
So we get the value as,
$\begin{aligned} & {{x}^{2}}-1=0 \\\ & \Rightarrow x=\pm 1 \\\ \end{aligned}$
$\begin{aligned} & x+1=0 \\\ & \Rightarrow x=-1 \\\ \end{aligned}$
$x=\pm 1,-1$…..$\left( 1 \right)$
Next we will find the value of below equation:
$\begin{aligned} & \left| {{x}^{3}}-4x-{{x}^{2}}+4 \right|=0 \\\ & \Rightarrow {{x}^{3}}-4x-{{x}^{2}}+4=0 \\\ & \Rightarrow x\left( {{x}^{2}}-4 \right)-1\left( {{x}^{2}}-4 \right)=0 \\\ & \Rightarrow \left( {{x}^{2}}-4 \right)\left( x-1 \right)=0 \\\ \end{aligned}$
So we get the values as:
$\begin{aligned} & \left( {{x}^{2}}-4 \right)=0 \\\ & \Rightarrow {{x}^{2}}=4 \\\ & \Rightarrow x=\sqrt{4} \\\ & \Rightarrow x=\pm 2 \\\ \end{aligned}$
$\begin{aligned} & x-1=0 \\\ & \Rightarrow x=1 \\\ \end{aligned}$
$x=\pm 2,1$…..$\left( 2 \right)$
So from equation (1) and (2) we get the common value as
$x=1$
Hence value of $x$ that satisfies the equation ${{\sin }^{2015}}y=\left| {{x}^{3}}+{{x}^{2}}-x-1 \right|+{{e}^{\left| {{x}^{3}}-4x-{{x}^{2}}+4 \right|}}+{{\tan }^{2}}2y+{{\cos }^{4}}y$ is 1.

Note: In this type of question we have to analyze what the question is trying to state. We shouldn’t do any calculation by solving both sides because there are two variables involved and we don’t know the value of any of them. Try to simplify the equation as much as possible before doing any calculation.