Question

Solve the expression and find the value of x.
\[\sin x+\cos =1+\sin x\cos x\].

Answer 1

Hint: Square the given expression on both sides. Apply basic trigonometric identities and solve to get the value of x.

Complete step-by-step answer:

Given to us is the expression,
\[\sin x+\cos =1+\sin x\cos x\]
Let us square both LHS and RHS.
\[{{\left( \sin x+\cos \right)}^{2}}={{\left( 1+\sin x\cos x \right)}^{2}}\]
We know \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
\[{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+{{\sin }^{2}}x{{\cos }^{2}}x+2\sin x\cos x\]
Let us cancel \[2\sin x\cos x\] from both the sides.
We know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\].
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x \\
 & 1=1+{{\sin }^{2}}x{{\cos }^{2}}x \\
\end{align}\]
Cancel out 1 from LHS and RHS. Hence we get,
\[{{\sin }^{2}}x{{\cos }^{2}}x=0\]
So we can write it as, \[\operatorname{sinx}=0\] or \[\cos x=0\].
\[x={{\sin }^{-1}}0\] and \[x={{\cos }^{-1}}0\]
From the trigonometric table we know \[\sin 0=0\] or \[\cos 90=0\].
Thus for \[\sin x=0,x=n\pi \]
and for \[\cos x=0,x=\dfrac{\left( 2n+1 \right)\pi }{2},n\in N\]
Thus we got the value of x as \[n\pi \] or \[\dfrac{\left( 2n+1 \right)\pi }{2},n\in N\].

Note: In the last step of the solution we must consider the condition that both the sin$x$ and cos$x$ can be zero, thereby we have to consider the angles of both because single alone happening zero will also make the complete multiplication zero.