Question
Question: : How do you solve \[4{\sin ^2}x + 5 = 6\]?...
: How do you solve 4sin2x+5=6?
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=6
Now transposing 5 to right hand side
⇒4sin2x=6−5
⇒4sin2x=1
Now, on cross-multiplication we get
⇒sin2x=41
Taking root on both sides
⇒sinx=±21
Therefore, there are two values of sinx
⇒sinx=21,−21
Now,sinx=21, when x=6πor65π
and sinx=−21, when x=−6πor−65π
as 65π=π−6π, we can summarize result, in general, as
Therefore, x=nπ±6π, 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 2π. In other words, after every 2π unit, the y-values repeat. If we need to find all possible solutions, then we must add 2πk, where k is an integer, to the initial solution.