Question

How do you solve $2 + \sec x = 0$ and find all solutions in the interval $0 < x < 360? $

Answer 1

Hint: In the given question we have $\sec x$ we will convert $\sec x$ into $\cos x$ by using the trigonometric identity $\sec x = \dfrac{1}{{\cos x}}$ and then we will find the value of $x$ which has to lie in the interval of $0 < x < 360$ .

Complete step by step solution:
As per given trigonometric equation,
We have,
$2 + \sec x = 0$
As we know that, $\sec x = \dfrac{1}{{\cos x}}$ (by trigonometric identity)
$ \Rightarrow 2 + \dfrac{1}{{\cos x}} = 0$
Now we will simplify the equation by shifting the constant number to one side and variable to another side. The resultant equation will be,
$ \Rightarrow \dfrac{1}{{\cos x}} = 0 - 2$
$ \Rightarrow \dfrac{1}{{\cos x}} = - 2$
Now we will Cross multiply, and the resultant equation will be
$ \Rightarrow - 2\cos x = 1$
Now we will shift $ - 2$ to the right side and equation will be,
$ \Rightarrow \cos x = \dfrac{{ - 1}}{2}$
Now in-order to find the value of$x$, we will transfer $\cos x$ on to the right side, and it will get converted into ${\cos ^{ - 1}}$ which is inverse of$\cos x$.
$ \Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$
Now we as know that ${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{2}{3}\pi $ from the trigonometric table, and the resultant equation will be,
$ \Rightarrow x = \dfrac{2}{3}\pi $
As we know that $\pi = 180^\circ $ we will place the value of $\pi $ in the above equation,
$ \Rightarrow x = \dfrac{2}{3} \times 180^\circ $
Now we will simplify the equation,
$ \Rightarrow x = \dfrac{{360^\circ }}{3}$
$ \Rightarrow x = 120^\circ $
And by given interval, $0 < x < 360^\circ $
The value of $'x'$between $0$ and $360^\circ $
$\therefore x = 360^\circ - 120^\circ $ $(120^\circ $ Is calculated value)
$\therefore x = 240^\circ $

Hence the value of $x$ are $120^\circ $ and $240^\circ $

Note: (1) While solving the above trigonometric equation apply the BODMAS first use division and then multiplication then addition and at last subtraction.
(2) As if $2 + \sec x = 0$ then if we change the place of any number from left to right side then its sign changes.
$' + '$ Will be converted into $' - '$
$' \times '$ Will be converted into $' \div '$