- #1
physior
- 182
- 1
hello
Let's solve:
(square root of 3)sin(x)-cos(x)=2
We know that R=square root of 3 + 1 = 2
We also know that tanφ=b/a= -1/sqrt3
The solution to tanφ=-1/sqrt3 has many solutions, for example, -30, 150, 330 degrees etc.
Which of these solutions do we accept? Or is it irrelevant which we will accept?
I mean, our equation will be this or that?
this: 2sin(x-30)=2 => sin(x-30)=1 => x-30 = 2kπ + 90 or x-30 = 2kπ + π - 90 => x=2kπ+120 or x=2kπ+π-60
that: 2sin(x+330)=2 => sin(x+330)=1 => x+330 = 2kπ + 90 or x+330 = 2kπ + π - 90 => x=2kπ-240 or x=2kπ+π-420
which of these solutions are acceptable?
thanks!
Let's solve:
(square root of 3)sin(x)-cos(x)=2
We know that R=square root of 3 + 1 = 2
We also know that tanφ=b/a= -1/sqrt3
The solution to tanφ=-1/sqrt3 has many solutions, for example, -30, 150, 330 degrees etc.
Which of these solutions do we accept? Or is it irrelevant which we will accept?
I mean, our equation will be this or that?
this: 2sin(x-30)=2 => sin(x-30)=1 => x-30 = 2kπ + 90 or x-30 = 2kπ + π - 90 => x=2kπ+120 or x=2kπ+π-60
that: 2sin(x+330)=2 => sin(x+330)=1 => x+330 = 2kπ + 90 or x+330 = 2kπ + π - 90 => x=2kπ-240 or x=2kπ+π-420
which of these solutions are acceptable?
thanks!