Question

How do you solve $\cos x - 2\sin x.\cos x = 0$ in the interval $\left[ {0,360} \right)$ ?

Answer 1

We are given with the expression and an interval also. Such that we are about to find the values of angle that satisfy the expression in the interval. So we will take the cos function common and then equate the brackets separately to zero just to solve the values of $x$.

Complete step by step solution:
We are given an expression $\cos x - 2\sin x.\cos x = 0$. We have to find the value of $x$ such that an interval is given $\left[ {0,360} \right)$. Taking cos function common,
$\cos x\left( {1 - 2\sin x} \right) = 0$
Thus either of the functions is zero. So we will equate them one by one to zero to find the value of $x$.
$\cos x = 0$
So we very well know that $\cos \dfrac{{n\pi }}{2} = 0$ but from the interval we will get $x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$.As the given interval is in degrees so we can write the value of angle as $x = {90^ \circ },{270^ \circ }$.

For the second function $\left( {1 - 2\sin x} \right) = 0$.
On solving we get,
$\sin x = \dfrac{1}{2}$
So we know that $\sin \dfrac{{n\pi }}{6} = \dfrac{1}{2}$ but from the interval we will get $x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$. As the given interval is in degrees so we can write the value of angle as $x = {30^ \circ },{150^ \circ }$.

Note: The interval is already given so we have a limit to write the value of the angle. The meaning or notations of the limit is it is open if having a parenthesis and it is closed if it is having a square bracket. There are limits given in the case above. So we took particular values. We checked for various values of $\dfrac{{n\pi }}{2}$ and $\dfrac{{n\pi }}{6}$. From them we came to get the values noted above because the other values are not correct for the value of the function.