Question

If \[x+\dfrac{1}{x}=2\cos \theta \], then \[x\] is equal to
A. \[\cos \theta +i\sin \theta \]
B. \[\cos \theta -i\sin \theta \]
C. \[\cos \theta \pm i\sin \theta \]
D. \[\sin \theta \pm i\cos \theta \]

Answer 1

Hint: If \[p(x)\] is a quadratic polynomial then \[p(x)=0\] is called a quadratic equation. The general form of the quadratic equation is \[a{{x}^{2}}+bx+c=0\]where \[a\ne 0\]. If \[p(x)\] is a quadratic equation then the zeroes of the polynomial \[p(x)\] are known as the roots of the equation. The roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] , \[a\ne 0\] can be found by using quadratic formula \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] , provided \[{{b}^{2}}-4ac\ge 0\]. The nature of the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\], \[a\ne 0\] depends upon the value of \[D={{b}^{2}}-4ac\] which is called the discriminant of the quadratic equation.

Complete step by step answer:
\[x+\dfrac{1}{x}=2\cos \theta \]
Solving the LHS( Left Hand Side) of the equation
\[\dfrac{{{x}^{2}}+1}{x}=2\cos \theta \]
Dividing both sides by \[x\] we get
\[{{x}^{2}}+1=2x\cos \theta \]
Transposing the whole term from RHS to LHS we get
\[{{x}^{2}}+1-2x\cos \theta =0\]
After rearranging the equation we get
\[{{x}^{2}}-2x\cos \theta +1=0\]
This equation is the in form of quadratic equation \[a{{x}^{2}}+bx+c=0\] where
\[a=1\]
\[\Rightarrow b=-2\cos \theta \]
\[\Rightarrow c=1\]
Applying the quadratic formula
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]

Substituting the values we get
\[x=\dfrac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}\]
Simplifying the equation we get
\[x=\dfrac{2\cos \theta \pm 2\sqrt{{{\cos }^{2}}\theta -1}}{2}\]
Using the trigonometric identity \[{{\cos }^{2}}\theta -1=-{{\sin }^{2}}\theta \] we get
\[x=\dfrac{2\cos \theta \pm 2\sqrt{-{{\sin }^{2}}\theta }}{2}\]
Here the \[D<0\] so the roots will be complex.
\[\sqrt{-{{\sin }^{2}}\theta }=i\sin \theta \]
Substituting this value we get
\[x=\dfrac{2\cos \theta \pm 2i\sin \theta }{2}\]
Taking \[2\] common from the numerator of RHS we get
\[x=\dfrac{2(\cos \theta \pm i\sin \theta )}{2}\]
Further solving we get
\[\therefore x=\cos \theta \pm i\sin \theta \]

Therefore, option \[C\] is the answer.

Note: \[i\] stands for iota. It is also called the imaginary part. A complex number has the real part and the imaginary part. The square root of the negative real number is called an imaginary quantity or imaginary number. If \[a,b\] are two real numbers then a number in the form \[a\pm ib\] is called a complex number.