I need to find the solutions in the interval [0, 2pi for:
2sinx + cosx = 0
Im trying to get it where I can split two things up and set them equal to zero but that can only be done when they are being multiplied to each other, but as you can see they are being added, so they cannot be separated.
how about,
2sinx = -cosx, now divide by cosx
2sinx/cosx = -1
sinx/cosx -1/2
tanx = -1/2
take inverse tan (+.5) to get 26.57º
the tangent is negative in quadrant II and IV
the angle is 180-26.57 and
360-26.57
Just noticed that you need radians, set your calculators on Radians, find the angle in standard position, then do
pi - angle and 2pi - angle.
To find the solutions for the equation 2sinx + cosx = 0 in the interval [0, 2pi], we need to use trigonometric identities and techniques.
First, let's try to simplify the equation:
2sinx + cosx = 0
We can rewrite this equation using the identity cosx = 1 - sin^2(x):
2sinx + 1 - sin^2(x) = 0
Rearranging the terms, we have:
sin^2(x) + 2sinx - 1 = 0
Now, we have a quadratic equation in terms of sinx. Let's solve it by factoring or using the quadratic formula.
To solve it by factoring, we need to find two values of sinx that satisfy the equation sin^2(x) + 2sinx - 1 = 0.
The equation cannot be factored simply, so we use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation sin^2(x) + 2sinx - 1 = 0, the coefficients are: a = 1, b = 2, c = -1.
Using the quadratic formula, we have:
sinx = (-2 ± sqrt(2^2 - 4(1)(-1))) / (2(1))
sinx = (-2 ± sqrt(4 + 4)) / 2
sinx = (-2 ± sqrt(8)) / 2
sinx = (-2 ± 2sqrt(2)) / 2
sinx = -1 ± sqrt(2)
Now, let's consider the interval [0, 2pi]. We know that sinx lies between -1 and 1.
Therefore, for sinx = -1 + sqrt(2):
Since -1 + sqrt(2) lies within the range of -1 and 1, we can solve for x by taking the inverse sine:
x = sin^(-1)(-1 + sqrt(2))
And for sinx = -1 - sqrt(2):
Since -1 - sqrt(2) does not lie within the range of -1 and 1, it does not have a valid solution in the interval [0, 2pi].
Hence, the solution for the equation 2sinx + cosx = 0 in the interval [0, 2pi] is:
x = sin^(-1)(-1 + sqrt(2))