Question

How do you solve $2\cos x - \sin x + 2\cos x\sin x = 1$ in the interval $0 \leqslant x \leqslant 2\pi $?

Answer 1

Hint: In this problem we have a trigonometric equation in some interval and we are asked to solve the trigonometric equation in the same interval. And to solve this given trigonometric equation we used to square the terms and also we may use some trigonometric identities. Here the value of $x$ lies in the given interval.

Formula used: $1 - {\sin ^2}x = {\cos ^2}x$
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$

Complete step-by-step solution:
The given equation is $2\cos x - \sin x + 2\cos x\sin x = 1$
Given equation can also be written as, $2\cos x - \sin x = 1 - 2\cos x\sin x$
Now, squaring on both sides, we get
${(2\cos x - \sin x)^2} = {(1 - 2\cos x\sin x)^2}$
Expanding the terms on both sides, by using the formula we get
$ \Rightarrow 4{\cos ^2}x + {\sin ^2}x - 4\cos x\sin x = 1 + 2{\cos ^2}x{\sin ^2}x - 4\cos x\sin x$
Now cancel the common terms on both sides, we get
$ \Rightarrow 4{\cos ^2}x + {\sin ^2}x = 1 + 2{\cos ^2}x{\sin ^2}x - - - - - (1)$
Here let us change all the terms into cosine and let us take $a = {\cos ^2}x$, also use the identity $1 - {\sin ^2}x = {\cos ^2}x$ in the equation (1).
Equation (1) becomes,
$ \Rightarrow 4a + 1 - a = 1 + 4a(1 - a)$
On simplify we get
$ \Rightarrow 4a + 1 - a = 1 + 4a - 4{a^2}$
Now cancel the common terms on both sides, we get
$ - a = - 4{a^2}$ Also cancel the minus sign on both sides
$ \Rightarrow a = 4{a^2}$
Let’s find the value of $a$
$ \Rightarrow a - 4{a^2} = 0$
Taking $a$ common on both sides we get
$ \Rightarrow a(1 - 4a) = 0$
On rewriting we get
$ \Rightarrow a = 0,1 - 4a = 0$
$ \Rightarrow a = 0,a = \dfrac{1}{4}$
But we have chosen $a = {\cos ^2}x$. Let’s substitute the value of $a$, we get
${\cos ^2}x = c = 0$ Or ${\cos ^2}x = c = \dfrac{1}{4}$
Take square root on both sides, we get
$\cos x = 0$ Or $\cos x = \pm \dfrac{1}{2}$
In the required range, $x = \dfrac{\pi }{2},x = \dfrac{{3\pi }}{2}$ from the first, $x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$ from the second
Let’s substitute these values in the given equation of left hand side, we get
When $x = \dfrac{\pi }{2}$,$2\cos \left( {\dfrac{\pi }{2}} \right) - \sin \left( {\dfrac{\pi }{2}} \right) + 2\cos \left( {\dfrac{\pi }{2}} \right)\sin \left( {\dfrac{\pi }{2}} \right) = - 1$ This is not correct.
When$x = \dfrac{\pi }{3}$, $2\cos \left( {\dfrac{\pi }{3}} \right) - \sin \left( {\dfrac{\pi }{3}} \right) + 2\cos \left( {\dfrac{\pi }{3}} \right)\sin \left( {\dfrac{\pi }{3}} \right) = 2\left( {\dfrac{1}{2}} \right) - \dfrac{{\sqrt 3 }}{2} + 2\left( {\dfrac{1}{2}} \right)\dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow 2\cos \left( {\dfrac{\pi }{3}} \right) - \sin \left( {\dfrac{\pi }{3}} \right) + 2\cos \left( {\dfrac{\pi }{3}} \right)\sin \left( {\dfrac{\pi }{3}} \right) = 1$. This is correct.
When$x = \dfrac{{2\pi }}{3}$, $2\cos \left( {\dfrac{{2\pi }}{3}} \right) - \sin \left( {\dfrac{{2\pi }}{3}} \right) + 2\cos \left( {\dfrac{{2\pi }}{3}} \right)\sin \left( {\dfrac{{2\pi }}{3}} \right) = 1 - \sqrt 3 $. This is not correct.
And this similarly for when $x = \dfrac{{4\pi }}{3}$
When $x = \dfrac{{5\pi }}{3}$, $2\cos \left( {\dfrac{{5\pi }}{3}} \right) - \sin \left( {\dfrac{{5\pi }}{3}} \right) + 2\cos \left( {\dfrac{{5\pi }}{3}} \right)\sin \left( {\dfrac{{5\pi }}{3}} \right) = 1$. This is correct.

Therefore, the required answer is $x = \dfrac{\pi }{3}$ or $x = \dfrac{{5\pi }}{3}$ that is, when we substitute these two values for $x$ in the left hand side given equation then the equation is satisfied.

Note: In this problem first we solved the given equation and we get $\cos x = 0$ or $\cos x = \pm \dfrac{1}{2}$.
If $x = \dfrac{\pi }{2}$ then $\cos x = 0$ also if $x = \dfrac{{3\pi }}{2}$ then $\cos x = 0$ .
Like the same way if $x = \dfrac{\pi }{3}$ then $\cos x = + \dfrac{1}{2}$, if $x = \dfrac{{2\pi }}{3}$ then $\cos x = - \dfrac{1}{2}$, if $x = \dfrac{{4\pi }}{3}$ then $\cos x = - \dfrac{1}{2}$, if $x = \dfrac{{5\pi }}{3}$ then $\cos x = \dfrac{1}{2}$.
These are things we found in the first part of this problem.
Then simply we substituted these $x$ values in the given equation to satisfy the equation and in that $x = \dfrac{\pi }{3},x = \dfrac{{5\pi }}{3}$ satisfied the given equation.