solve for 0=/<x<2pi
6sin^2x + sinx - 1 = 0
Ahh, forgot how to factor.
3cos^2 - 2cos - 1 = 0
(3cosx + 1 )(cosx - 1)
cosx= -1/3, 1
x=0, and .....?
You seem to be doing two different problems.
The first one can be factored to
(3 sinx -1)(2sin x + 1) = 0
sin x = 1/3 or -1/2
x = 19.47, 160.53, 150, 330 (degrees)
Express them in radians if necessary
19.47 degrees is arcsin (1/3) = 0.3398 radians
solve for 0=/<x<2pi
6sin^2x + sinx - 1 = 0
Ahh, forgot how to factor.
6 y^2 + y - 1 = 0
(3 y-1)(2y+1) = 0
y = sin x = 1/3
y = sin x =-1/2
3cos^2 - 2cos - 1 = 0
(3cosx + 1 )(cosx - 1)
cosx= -1/3, 1
x=0, and .....? 2 pi
x = 1.91 radians and 4.37 radians
so x = 0.3398, 2.807, 7pi/6 and 11pi/6
thanks!
To solve the equation 6sin^2x + sinx - 1 = 0, we can use the quadratic formula or factorize it. In this case, let's factorize it.
Step 1: Rewrite the equation in a quadratic form by rearranging the terms:
6sin^2x + sinx - 1 = 0
Step 2: Factorize the quadratic equation:
(2sinx - 1)(3sinx + 1) = 0
Now, we have two cases to consider:
Case 1: 2sinx - 1 = 0
Solving this equation gives us:
2sinx = 1
sinx = 1/2
To find the values of x within the given range of 0 ≤ x < 2pi, we can refer to the unit circle or remember the special angles:
x = pi/6 (30 degrees) and x = 5pi/6 (150 degrees)
Case 2: 3sinx + 1 = 0
Solving this equation gives us:
3sinx = -1
sinx = -1/3
Again, referring to the unit circle or remembering the special angles, we can find the values of x within the given range:
x = 7pi/6 (210 degrees) and x = 11pi/6 (330 degrees)
So, the solutions for the equation 6sin^2x + sinx - 1 = 0 within the range 0 ≤ x < 2pi are:
x = pi/6, 5pi/6, 7pi/6, 11pi/6