Find real \[\theta \] for which complex number \[\dfrac{{1 + i\cos \theta }}{{1 - 2i\cos \theta }}\] is purely real.

Hint: Here we rationalize the given complex number by multiplying both numerator and denominator with the conjugate of the denominator of the given complex number. Then using the fact that the complex number is purely real, we equate the imaginary part to zero which gives us the required value of \[\theta \].
* In a complex number \[z = x + iy\], the real part is x and the imaginary part is y.
* If \[z = x + iy\]is a complex number, then conjugate of this complex number is \[\overline z = x - iy\]
* The value of \[i = \sqrt { - 1} \]. On squaring both sides it gives \[{i^2} = - 1\].

Complete step-by-step answer:
We are given the complex number as \[\dfrac{{1 + i\cos \theta }}{{1 - 2i\cos \theta }}\] … (1)
We take the denominator of the complex number \[1 - 2i\cos \theta \].
Now we write the conjugate of the denominator.
\[\overline {1 - 2i\cos \theta } = 1 + 2i\cos \theta \]
Now we multiply the numerator and denominator of complex number in equation (1) by \[1 + 2i\cos \theta \]
\[ \Rightarrow \dfrac{{1 + i\cos \theta }}{{1 - 2i\cos \theta }} \times \dfrac{{1 + 2i\cos \theta }}{{1 + 2i\cos \theta }}\]
\[ \Rightarrow \dfrac{{(1 + i\cos \theta ) \times (1 + 2i\cos \theta )}}{{(1 - 2i\cos \theta ) \times (1 + 2i\cos \theta )}}\]
Use the formula \[(a + b)(a - b) = {a^2} - {b^2}\] in the denominator.
\[ \Rightarrow \dfrac{{1 \times (1 + 2i\cos \theta ) + i\cos \theta \times (1 + 2i\cos \theta )}}{{{{(1)}^2} - {{(2i\cos \theta )}^2}}}\]
\[ \Rightarrow \dfrac{{1 + 2i\cos \theta + i\cos \theta + 2{i^2}{{\cos }^2}\theta }}{{1 - 4{i^2}{{\cos }^2}\theta }}\]
Substitute the value of \[{i^2} = - 1\] in both numerator and denominator
\[ \Rightarrow \dfrac{{1 + 2i\cos \theta + i\cos \theta + 2( - 1){{\cos }^2}\theta }}{{1 - 4( - 1){{\cos }^2}\theta }}\]
\[ \Rightarrow \dfrac{{1 + 3i\cos \theta - 2{{\cos }^2}\theta }}{{1 + 4{{\cos }^2}\theta }}\]
Combine and write the real part and imaginary part separately.
\[ \Rightarrow \dfrac{{(1 - 2{{\cos }^2}\theta ) + 3i\cos \theta }}{{1 + 4{{\cos }^2}\theta }}\]
\[ \Rightarrow \dfrac{{1 - 2{{\cos }^2}\theta }}{{1 + 4{{\cos }^2}\theta }} + i\dfrac{{3\cos \theta }}{{1 + 4{{\cos }^2}\theta }}\]
Comparing the complex number with the general complex number \[z = x + iy\], where x is real part and y is imaginary part.
Here real part is \[\dfrac{{1 - 2{{\cos }^2}\theta }}{{1 + 4{{\cos }^2}\theta }}\] and imaginary part is \[\dfrac{{3\cos \theta }}{{1 + 4{{\cos }^2}\theta }}\]
For the complex number to be purely real, the imaginary part has to be zero.
\[ \Rightarrow \dfrac{{3\cos \theta }}{{1 + 4{{\cos }^2}\theta }} = 0\]
Cross multiply the terms in the denominator of LHS to RHS of the equation.
\[ \Rightarrow 3\cos \theta = 0\]
Divide both sides of the equation by 3
\[ \Rightarrow \dfrac{{3\cos \theta }}{3} = \dfrac{0}{3}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \cos \theta = 0\]
We know \[\cos {90^ \circ } = 0\], therefore, substitute the value in RHS
\[ \Rightarrow \cos \theta = \cos {90^ \circ }\]
Comparing the angles on both sides we get
\[\theta = {90^ \circ }\].

Note: Students many times make mistakes when they try to separate the given complex number without even rationalizing it which is wrong, because a standard form of complex number does not have i in its denominator. Also, some students write the value of the imaginary part with ‘i’ along it which should be taken care of, as the imaginary part is the value except i.