find the exact solutions
2cos^2x+3sinx=0
the way it stands, that is a "nasty" question.
Are you sure the second term isn't 2sin(2x) ?
no, its as i wrote it.
then it's got me stymied, I must be missing something rather obvious, sorry!
thanks for trying!
2cos^2x+3sinx=0
2-2sin^2x + 3 sinx=0
sin^2 x - 3/2 sin x - 1=0
quadratic equation..
sinx= (3/2 +-sqrt(9/4 +4) /2
sinx= 3/4 +- 5/4
sin x= -1/2 works.
check me.
thank you!
To find the exact solutions of the equation 2cos^2x + 3sinx = 0, we can simplify it algebraically.
First, we can expand the expression by using the identity cos^2x = 1 - sin^2x:
2(1 - sin^2x) + 3sinx = 0
Next, we simplify this equation:
2 - 2sin^2x + 3sinx = 0
To further simplify, we rearrange terms and collect like terms:
-2sin^2x + 3sinx + 2 = 0
Now, we have a quadratic equation in terms of sinx. Let's rewrite it as follows:
sin^2x - (3/2)sinx - 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
sinx = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our case, a = 1, b = -(3/2), and c = -1. Plugging these values into the quadratic formula, we get:
sinx = (-(3/2) ± sqrt((3/2)^2 - 4(1)(-1))) / (2(1))
Simplifying further:
sinx = (-(3/2) ± sqrt(9/4 + 4)) / 2
sinx = (-(3/2) ± sqrt(9/4 + 16/4)) / 2
sinx = (-(3/2) ± sqrt(25/4)) / 2
sinx = (-(3/2) ± 5/2) / 2
This gives us two possible solutions:
1) sinx = (-(3/2) + 5/2) / 2 = 1/2
2) sinx = (-(3/2) - 5/2) / 2 = -2
Since we are looking for solutions in the range of -1 to 1, the second solution (-2) is not valid. Therefore, the only exact solution to the equation 2cos^2x + 3sinx = 0 is sinx = 1/2.
You can verify this solution by substituting sinx = 1/2 back into the original equation to check that it satisfies the equation.