the problem is
2cos^2x + sinx-1=0
the directions are to "use an identity to solve each equation on the interval [0,2pi).
This is what i've done so far:
2cos^2x+sinx-1=0
2cos^2x-1+sinx=0
cos2x + sinx =0
1 - 2sin^2x + sinx = 0
-2sin^2x+sinx-1=0
2sin^2x-sinx+1=0
I think i'm supposed to factor it next but i'm not sure?
(2sinx-?)(sinx-?)
would doing that step be correct?
although, after looking at this problem i'm still not sure why i had to change the original problem? I've just been following the notes that were given to me by my teacher and that's what she has done but why do i have to change the original problem it seems like just a bunch of extra work
2cos^2x+sinx-1=0
replace cos^2x by 1-sin^2x
2(1-sin^2x) + sinx -1 = 0
2-2sin^2x +sinx -1 = 0
-2sin^2x+sinx+1=0
2sin^2x - sinx - 1=0
(2sinx + 1)(sinx -1)=0
we will get
sinx = -1/2 and 1
carrying on where Dena left off....
sinx = -1/2 (angle in standard postition = 30º)
x must be in quadrant III or IV
x = 180+30 =210 or
x = 360-30 = 330
if sinx = 1
x = 90
so in degrees x = 90, 210 or 330
or in radians
x = pi/2, 7pi/6 or 11pi/6
Yes, you are on the right track. The next step is to factor the quadratic equation 2sin^2x - sinx + 1 = 0. However, to factor it correctly, we need to find two numbers that multiply to give 2 and add up to -1.
To find the factors, we can use the product-sum method:
- First, multiply the coefficient of sin^2x (which is 2) by the constant term (which is 1).
2 * 1 = 2
- Now, we need to find two numbers that multiply to give 2 and add up to -1. After trying different combinations, we can determine that the numbers are -2 and -1.
-2 + (-1) = -3
Now, we can rewrite the quadratic equation using these numbers:
2sin^2x - 2sinx - sinx + 1 = 0
Next, we can factor the expression by grouping:
(2sin^2x - 2sinx) - (sinx - 1) = 0
Now, factor out the common factors from each group:
2sinx(sin(x) - 1) - 1(sin(x) - 1) = 0
Notice that (sin(x) - 1) appears as a common factor in both terms. We can now factor it out:
(2sinx - 1)(sin(x) - 1) = 0
So, the factored form of the equation is (2sinx - 1)(sin(x) - 1) = 0.
Now, to solve for x, we set each factor equal to zero and solve for x:
2sinx - 1 = 0 => sinx = 1/2
sin(x) - 1 = 0 => sinx = 1
To find the solutions on the interval [0, 2π), we need to determine the values of x that satisfy these equations. The sine function is positive for angles in the first and second quadrant, so on the interval [0, 2π), the solutions are:
x = π/6, 5π/6, and π.