Help find the general solution to:sin^2x-3cosx+sinxcosx-3sinx=0
sin^2x-3cosx+sinxcosx-3sinx=0
sin^2x - 3sinx + sinxcosx - 3cosx = 0
sinx(sinx - 3) + cosx(sinx - 3) = 0
(sinx - 3)(sinx + cosx) = 0
sinx = 3 , which is not possible
or
sinx = -cosx
sinx/cosx = -1
tanx = -1
x = 135° or 225° or in radians, 3π/4 , 5π/4
the period of tanx = 180° or π radians
general solution: = 3π/4 + kπ
or
x = 5π/4 + kπ , where k is an integer
To find the general solution to the given equation: sin^2x - 3cosx + sinxcosx - 3sinx = 0, we can manipulate the equation and simplify it.
First, let's group similar terms:
sin^2x + sinxcosx - 3sinx - 3cosx = 0
Next, let's factor out sinx from the first two terms and factor out -3 from the last two terms:
sinx(sinx + cosx) - 3(sin x + cos x) = 0
Now, notice that (sinx + cosx) appears in both terms. We can factor it out:
(sin x + cos x)(sin x - 3) = 0
Now, we have two separate equations:
1. sin x + cos x = 0
2. sin x - 3 = 0
Let's solve each equation separately:
1. sin x + cos x = 0:
We can rewrite this equation as:
cos x = -sin x
Divide both sides by cos x:
1 = -tan x
Taking the inverse tangent (arctan) of both sides, we get:
x = arctan(-1) + nπ, where n is an integer.
Simplifying further, we have:
x = -π/4 + nπ
2. sin x - 3 = 0:
Adding 3 to both sides, we get:
sin x = 3
However, there are no solutions to this equation, as the range of sin x is -1 to 1.
Hence, this equation has no solutions.
Therefore, the general solution to the given equation sin^2x - 3cosx + sinxcosx - 3sinx = 0 is:
x = -π/4 + nπ, where n is an integer.