Solveeit Logo

Question

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

: How do you solve 4sin2x+5=64{\sin ^2}x + 5 = 6?

Explanation

Solution

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 4sin2x+5=64{\sin ^2}x + 5 = 6
Now transposing 55 to right hand side
4sin2x=65\Rightarrow 4{\sin ^2}x = 6 - 5
4sin2x=1\Rightarrow 4{\sin ^2}x = 1
Now, on cross-multiplication we get
sin2x=14\Rightarrow {\sin ^2}x = \dfrac{1}{4}
Taking root on both sides
sinx=±12\Rightarrow \sin x = \pm \dfrac{1}{2}
Therefore, there are two values of sinx\sin x
sinx=12,12\Rightarrow \sin x = \dfrac{1}{2}, - \dfrac{1}{2}
Now,sinx=12\sin x = \dfrac{1}{2}, when x=π6  or  5π6x = \dfrac{\pi }{6}\;or\;\dfrac{{5\pi }}{6}
and sinx=12\sin x = - \dfrac{1}{2}, when x=π6  or  5π6x = - \dfrac{\pi }{6}\;or - \;\dfrac{{5\pi }}{6}
as 5π6=ππ6\dfrac{{5\pi }}{6} = \pi - \dfrac{\pi }{6}, we can summarize result, in general, as
Therefore, x=nπ±π6x = 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 2π2\pi . In other words, after every 2π2\pi unit, the y-values repeat. If we need to find all possible solutions, then we must add 2πk2\pi k, where k is an integer, to the initial solution.