Help find the general solution to:sin^2x-3cosx+sinxcosx-3sinx=0
To find the general solution of the given equation, let's start by taking a closer look at it:
sin^2(x) - 3cos(x) + sin(x)cos(x) - 3sin(x) = 0
To simplify this equation, we can rearrange the terms:
sin^2(x) + sin(x)cos(x) - 3cos(x) - 3sin(x) = 0
Next, let's rewrite sin(x)cos(x) as 0.5sin(2x). This will help us simplify the equation further:
sin^2(x) + 0.5sin(2x) - 3cos(x) - 3sin(x) = 0
Now, let's group the terms with sin(x) and cos(x):
(sin^2(x) - 3sin(x)) + (0.5sin(2x) - 3cos(x)) = 0
Now, let's factor out sin(x) from the first group and cos(x) from the second group:
sin(x)(sin(x) - 3) + 0.5(sin(2x) - 6cos(x)) = 0
Using the double angle identity sin(2x) = 2sin(x)cos(x), we can simplify further:
sin(x)(sin(x) - 3) + 0.5(2sin(x)cos(x) - 6cos(x)) = 0
sin(x)(sin(x) - 3) + sin(x)cos(x) - 3cos(x) = 0
Now, let's regroup the terms again:
sin(x)(sin(x) + cos(x) - 3) - 3cos(x) = 0
At this point, we can see that we have two factors: sin(x) and (sin(x) + cos(x) - 3). For the equation to be true, one or both of these factors must equal zero.
Setting sin(x) = 0, we have x = 0, π, 2π, ...
Setting sin(x) + cos(x) - 3 = 0 requires a bit more work. We can rewrite it as cos(x) = 3 - sin(x). Squaring both sides, we get:
cos^2(x) = (3 - sin(x))^2
1 - sin^2(x) = 9 - 6sin(x) + sin^2(x)
2sin^2(x) - 6sin(x) + 8 = 0
Divide by 2:
sin^2(x) - 3sin(x) + 4 = 0
Now, we have a quadratic equation in terms of sin(x). We can solve it using the quadratic formula:
sin(x) = (-(-3) ± √((-3)^2 - 4(1)(4))) / (2(1))
sin(x) = (3 ± √(9 - 16)) / 2
sin(x) = (3 ± √(-7)) / 2
Since there is no real square root of -7, there are no solutions for sin(x) in this case.
So, the general solution to the given equation is:
x = 0, π, 2π, ...