Question

How do you solve \[\cos 2x=\sin x\] on the interval \[0\le x\le 2\pi \] ?

Answer 1

Hint: To solve the trigonometric equations which are basically the linear equations, we need to bring all the useful concepts together. We use trigonometric identities, reference angles, and the values of functions as well as algebra in solving these equations. Sometimes, we need to use substitution and factorization methods too. We should also take care of the interval in which the function would yield positive or negative values.

Complete step by step solution:
We have to solve the equation \[cos\text{ }2x=sin\text{ }x\] on the interval \[0\le x\le 2\pi \]
That is we have to find the value of x lying in the given interval such that the equation satisfies that value .
\[\begin{align}
  & cos\text{ }2x=sin\text{ }x \\
 & cos\text{ }2x-sin\text{ }x=0 \\
\end{align}\]
Now , we will trigonometric identity to simplify \[cos\text{ }2x\]
$cos2x=1-2si{{n}^{2}}x$
Using this , we get
\[\begin{align}
  & 1-2si{{n}^{2}}x-sinx=0 \\
 & 2si{{n}^{2}}x+sinx-1=0 \\
\end{align}\]
Now we will use substitution for \[sinx\]
Let \[sinx=y\]
Then the equation becomes
$2{{y}^{2}}+y-1=0$
Now we can solve this equation like any other quadratic equation
$2{{y}^{2}}+2y-y-1=0$
$2y(y+1)-1(y+1)=0$
$(2y-1)(y+1)=0$
$2y-1=0$ or $y+1=0$
$y=\dfrac{1}{2}$ or $y=-1$
This implies
$\sin x=\dfrac{1}{2}$ or $\sin x=-1$
We know that sin is positive in first and second quadrant while negative in third and fourth quadrant
So when $\sin x=\dfrac{1}{2}$ , the value of $x$ lies in first and second quadrant
We know $\sin x=\dfrac{1}{2}$ when \[x=\pi /6\] where \[\pi /6\] lies in first quadrant,
and \[x=\pi -\pi /6=\dfrac{5\pi }{6}\] which lies in second quadrant
When $\sin x=-1$, the value of $x$ lies in the third and fourth quadrant .
For $x$ in \[\left[ 0,2\pi \right]\], we get
\[\begin{array}{*{35}{l}}
   x=\pi +\dfrac{\pi }{2} \\
   x=\dfrac{3\pi }{2} \\
\end{array}\]

Thus the solution of \[cos\text{ }2x=sin\text{ }x\] in the interval \[\left[ 0,2\pi \right]\] is $\dfrac{\pi }{6},\dfrac{5\pi }{6},\dfrac{3\pi }{2}$

Note:
There are many values of $x$ for which the equation may satisfy but we have to find the value for the interval \[\left[ 0,2\pi \right]\]. So, we will not include any more value.