Can you please help me solve this trig equation for 0 ≤ x ≤ 2π?
9sin^2(x)-6cos(x)-10=0
Thank you!
Turn it into a quadratic equation for u = cos x.
9(1 - cos^2x)-6 cos x - 10 = 0
-9u^2 -6u -1 = 0
(3u +1)^2 = 0
cos x = -1/3
There will be solutions in the second and third quadrant
Thanks.
Am I supposed to use the inverse trig function to find the angle?
To solve the trigonometric equation 9sin^2(x) - 6cos(x) - 10 = 0 for values of x between 0 and 2π, we can follow these steps:
Step 1: Rearrange the equation
Let's rearrange the equation to have one trigonometric function in terms of the other. Since sin^2(x) = 1 - cos^2(x), we can substitute it in the equation:
9(1 - cos^2(x)) - 6cos(x) - 10 = 0
Step 2: Simplify the equation
Distribute the 9 through the parentheses:
9 - 9cos^2(x) - 6cos(x) - 10 = 0
Combine like terms:
-9cos^2(x) - 6cos(x) - 1 = 0
Step 3: Solve the quadratic equation
Now we have a quadratic equation in terms of cos(x). Let's substitute cos(x) with a new variable, say t, to make it easier to solve:
-9t^2 - 6t - 1 = 0
We can solve this quadratic equation using the quadratic formula:
t = [-b ± sqrt(b^2 - 4ac)] / (2a)
In this case, a = -9, b = -6, and c = -1. Plugging these values into the formula:
t = [-(-6) ± sqrt((-6)^2 - 4(-9)(-1))] / (2(-9))
t = [6 ± sqrt(36 - 36)] / (-18)
t = [6 ± 0] / (-18)
t = 0 / (-18) or 12 / (-18)
Simplifying, we get:
t = 0 or t = -2/3
Step 4: Solve for x
To solve for x, we use the inverse cosine function (arccos) since cos(x) = t:
For t = 0:
cos(x) = 0
x = π/2 or x = 3π/2 (according to the given range)
For t = -2/3:
cos(x) = -2/3
x = arccos(-2/3)
Using a calculator, we can find the approximate value of arccos(-2/3):
x ≈ 2.3005
So, the solutions for the trigonometric equation 9sin^2(x) - 6cos(x) - 10 = 0, for 0 ≤ x ≤ 2π, are:
x = π/2, 3π/2, and approximately 2.3005.