Question

: How do you solve \[4{\sin ^2}x + 5 = 6\]?

Answer 1

Hint: In this question, first we have to simplify this question and convert it into a form so that we can take the root of it. After that we will find the multiple solutions it posses as it contains both positive and negative values of sine function.

Complete step by step answer:
In the above question, we have given that \[4{\sin ^2}x + 5 = 6\]
Now transposing $5$ to right hand side
$ \Rightarrow 4{\sin ^2}x = 6 - 5$
$ \Rightarrow 4{\sin ^2}x = 1$
Now, on cross-multiplication we get
$ \Rightarrow {\sin ^2}x = \dfrac{1}{4}$
Taking root on both sides
$ \Rightarrow \sin x = \pm \dfrac{1}{2}$
Therefore, there are two values of $\sin x$
$ \Rightarrow \sin x = \dfrac{1}{2}, - \dfrac{1}{2}$
Now,$\sin x = \dfrac{1}{2}$, when \[x = \dfrac{\pi }{6}\;or\;\dfrac{{5\pi }}{6}\]
and $\sin x = - \dfrac{1}{2}$, when \[x = - \dfrac{\pi }{6}\;or - \;\dfrac{{5\pi }}{6}\]
as $\dfrac{{5\pi }}{6} = \pi - \dfrac{\pi }{6}$, we can summarize result, in general, as
Therefore, \[x = n\pi \pm \dfrac{\pi }{6}\], where n is an integer.

Note:
Trigonometric equations may have an infinite number of solutions. The period of both the sine function and the cosine function is $2\pi $. In other words, after every $2\pi $ unit, the y-values repeat. If we need to find all possible solutions, then we must add $2\pi k$, where k is an integer, to the initial solution.