Question

Solve \[\sin x+\cos x=1+\sin x\cos x\] , then the solution of x is \[2n\pi \pm \dfrac{\pi }{4}\] . If true then enter 1 and if false then enter 0.

Answer 1

Hint: First of all, square both sides of the equation \[\sin x+\cos x=1+\sin x\cos x\] . Now, use the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] and simplify the equation \[({{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x)\] . Now, we have to get the solution of \[\sin x=0\] and \[\cos x=0\] . We know that \[\sin 0=0\] , \[\sin \pi =0\] , and \[\sin 2\pi =0\] . So, \[\sin x=\sin n\pi \] . We know that \[\cos \dfrac{\pi }{2}=0\] , \[\cos \left( -\dfrac{\pi }{2} \right)=0\] , and \[\cos \left( \dfrac{3\pi }{2} \right)=0\] . So, \[\cos x=\cos \left( \dfrac{3\pi }{2} \right)\] . Now, solve it further and get the value of x.

Complete step-by-step answer:
According to the question, our given expression is \[\sin x+\cos x=1+\sin x\cos x\] ……………(1)
We have to get the solutions of x.
Now, squaring on both LHS and RHS of the equation (1), we get
\[\sin x+\cos x=1+\sin x\cos x\]
\[\Rightarrow {{(\sin x+\cos x)}^{2}}={{(1+\sin x\cos x)}^{2}}\] ……………………………(2)
We know the formula, \[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] . We can expand equation (2) with the help of the formula.
Using the formula and expanding equation (2), we get
\[\Rightarrow {{(\sin x+\cos x)}^{2}}={{(1+\sin x\cos x)}^{2}}\]
\[\begin{align}
  & \Rightarrow {{(\sin x+\cos x)}^{2}}={{(1+\sin x\cos x)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+{{\sin }^{2}}x{{\cos }^{2}}x+2\sin x\cos x \\
\end{align}\]
\[\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x\] ………………………(3)
We know the identity, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] ……………(4)
Using equation (4) and replacing \[({{\sin }^{2}}x+{{\cos }^{2}}x)\] by 1 in equation (3), we get
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x \\
 & \Rightarrow 1=1+{{\sin }^{2}}x{{\cos }^{2}}x \\
\end{align}\]
\[\Rightarrow 0={{\sin }^{2}}x{{\cos }^{2}}x\]
\[\Rightarrow 0={{\sin }^{2}}x{{\cos }^{2}}x\]
\[\Rightarrow 0=\sin x\cos x\]
So, either \[\sin x=0\] or \[\cos x=0\] ………………..(5)
Let us solve \[\sin x=0\] .
We know that, \[\sin 0=0\] …………….(6)
From equation (5) and equation (6), we get
\[\begin{align}
  & \Rightarrow \sin x=\sin 0 \\
 & \Rightarrow x=0 \\
\end{align}\]
We know that, \[\sin \pi =0\] ……………...(7)
From equation (5) and equation (7), we get
\[\begin{align}
  & \Rightarrow \sin x=\sin \pi \\
 & \Rightarrow x=\pi \\
\end{align}\]
We know that, \[\sin 2\pi =0\] …………...(8)
From equation (5) and equation (8), we get
\[\begin{align}
  & \Rightarrow \sin x=\sin 2\pi \\
 & \Rightarrow x=2\pi \\
\end{align}\]
We can see that x is a multiple of \[\pi \] .
So, \[x=n\pi \] ……………………….(9)
Let us solve \[\cos x=0\] .
We know that, \[\cos \dfrac{\pi }{2}=0\] …………….(10)
From equation (5) and equation (10), we get
\[\begin{align}
  & \Rightarrow \cos x=\cos \dfrac{\pi }{2} \\
 & \Rightarrow x=\dfrac{\pi }{2} \\
\end{align}\]
We know that, \[\cos \dfrac{3\pi }{2}=0\] ……………...(11)
From equation (5) and equation (11), we get
\[\begin{align}
  & \Rightarrow \cos x=\cos \dfrac{3\pi }{2} \\
 & \Rightarrow x=\dfrac{3\pi }{2} \\
\end{align}\]
We know that, \[\cos \left( -\dfrac{\pi }{2} \right)=0\] …………...(12)
From equation (5) and equation (12), we get
\[\begin{align}
  & \Rightarrow \cos x=\cos \left( -\dfrac{\pi }{2} \right) \\
 & \Rightarrow x=\dfrac{-\pi }{2} \\
\end{align}\]
Now, we have values of x which are \[-\dfrac{\pi }{2},\dfrac{\pi }{2},and\dfrac{3\pi }{2}\] .
So,
\[x=n\pi \pm \dfrac{\pi }{2}\] ……………………(13)
From equation (9) and equation (13), we have the solutions for x which are \[x=n\pi \] and \[x=n\pi \pm \dfrac{\pi }{2}\] .
So, \[x=2n\pi \pm \dfrac{\pi }{4}\] is not the solution of x.
It is given that we have to enter 1 if true and if false then we have to enter 0.
Thus, we have to enter 0.

Note: In this question, one may think to use the formula \[\sin 2x=2\sin x\cos x\] to solve the equation
\[({{\sin }^{2}}x{{\cos }^{2}}x=0)\] . Using the formula, \[\sin 2x=2\sin x\cos x\] we can replace \[{{\sin }^{2}}x{{\cos }^{2}}x\] by \[\dfrac{{{\sin }^{2}}2x}{4}\] .
\[\begin{align}
  & \dfrac{{{\sin }^{2}}2x}{4}=0 \\
 & \Rightarrow \sin 2x=0 \\
\end{align}\]
Now, solving \[\sin 2x=0\] ,
\[\begin{align}
  & \sin 2x=\sin n\pi \\
 & \Rightarrow 2x=n\pi \\
 & \Rightarrow x=\dfrac{n\pi }{2} \\
\end{align}\]
Here, we have got only \[x=\dfrac{n\pi }{2}\] . Using this formula makes us lose the solution of \[\cos x=0\] . So, we don’t have to use this formula here.