find all solutions on the inverval [0, 2pi) for sin2x=1 how do i solve this? Actually, I need to understand this as if I were a 2 year old. I am taking Trig and I was doing well until I came to these types of math problems.
\[\sin 2x = 1 \rightarrow 2x = n \pi + (-1)^{n}(\frac{ \pi }{ 2 })\]\[\rightarrow x = n \frac{ \pi }{ 2 } + (-1)^n (\frac{ \pi }{ 4 })\] If n = 0, x = pi/4. If n = 1, x = pi/4 again. If n = 2, x = pi + pi/4 = 5pi/4.
ooooo...so this is like solving a quadratic?
The general solution of sin(fn) is \[2x = n \times \pi + (-1)^{n} \times \theta \] n-is any integer theta - one of the solution here one of the solution is pi/4. hence getthe others.
If you expect yourself to ba a 2 year old and try to attempt these problems, I am sorry to say I can't help.
Thank you both for your reply. I just need an explanation on how to get to the solutions, like the steps, thats what I dont understand. And the 2-year old comparison was just to let people that wish to help to just explain it in simple terms. your comment, Rishi, is out of place.
Try studying trignometric equations from any good book. You will get a good idea of solving these problems.
thanks koushik, I do have the book and online resources but I get comfort in people actually explaining it, a small setback when learning online. I appreciate everybody's response.
Actually, you need a knowledge of trigonometry. But without any knowledge of it, it's impossible to solve these questions. This is the very basic trigonometric equation. Let me tell you a general technique. First, solve for x for a positive acute angle. In this case, sin2x=1 gives x = pi/4. Now, you should know that if x = 3pi/4 then also you would get sin2x=1. Likewise, there are different solutions of x but you can't go on guessing them, right? So, we need a general formula which can give all possible values of x.
Yeah, so we need to first know all the basic equations and their general solutions. Once you have them in your memory, you can reduce the equation to the known one and apply the known solution.
again, Thanks a lot, i know how the graphs look like but Rishi's explanation just now made me see the solution to the problem!
Good,happy that you could get the solution.
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