Find all values of theta in the interval 0 <= theta < 360 that satisfy the equation 3cos(2theta)=7cos(theta). Express answers to the nearest ten minutes.
Cos2T= sin^2T-Cos^2T=1-2cos^2T
3-6cos^2T=7cosT
let u=cosT
6u^2+7u-3=0
(3u-1)(2u+3)=0
u= 1/3 u= -2/3
so what angles are those?
I must respectfully disagree with bob
cos 2T = cos^2 T - sin^2 T = 2cos^2 T - 1
bob had it backwards, time for a strong coffee.
To find all the values of theta that satisfy the equation 3cos(2theta) = 7cos(theta) in the given interval, we can use some trigonometric identities and solve for theta step by step.
Let's start by simplifying the equation using the double angle identity for cosine: cos(2theta) = 2cos^2(theta) - 1. Substituting this back into the equation, we have:
3(2cos^2(theta) - 1) = 7cos(theta)
Simplifying further:
6cos^2(theta) - 3 = 7cos(theta)
Rearranging the equation:
6cos^2(theta) - 7cos(theta) - 3 = 0
Now we have a quadratic equation in terms of cos(theta). We can solve it by factoring, completing the square, or using the quadratic formula. In this case, factoring would be the easiest method:
(2cos(theta) + 1)(3cos(theta) - 3) = 0
Setting each factor equal to zero:
2cos(theta) + 1 = 0 or 3cos(theta) - 3 = 0
Solving for cos(theta) in each equation:
2cos(theta) = -1 or 3cos(theta) = 3
cos(theta) = -1/2 or cos(theta) = 1
Now we have found two values of cos(theta) that satisfy the equation: -1/2 and 1.
To find the values of theta, we need to find the angles whose cosine is equal to -1/2 or 1. Using a calculator or reference table, we can find the corresponding values of theta in the interval 0 <= theta < 360.
For cos(theta) = -1/2, the reference angle is 60 degrees (or pi/3 radians). Since cos(theta) is negative in the second and third quadrants, we can subtract the reference angle from 180 to find the solutions:
theta = 180 - 60 = 120 degrees (or 2pi - pi/3 = 5pi/3 radians)
theta = 180 + 60 = 240 degrees (or 2pi + pi/3 = 7pi/3 radians)
For cos(theta) = 1, the reference angle is 0 degrees (or 0 radians). Since cos(theta) is positive in the first and fourth quadrants, the solutions are:
theta = 0 degrees (or 0 radians)
theta = 360 degrees (or 2pi radians)
So the values of theta that satisfy the equation 3cos(2theta) = 7cos(theta) in the interval 0 <= theta < 360 are:
theta = 0 degrees, 120 degrees, 240 degrees, 360 degrees (or 0 radians, 5pi/3 radians, 7pi/3 radians, 2pi radians).
3cos(2Ø)=7cos(Ø)
3(2cos^2 Ø - 1) = 7cosØ
6cos^ Ø - 7cosØ - 3 = 0
(3cosØ+1)(2cosØ-3) = 0
cosØ = -1/3 or cosØ = 3/2, the last is not possible
cosØ = -1/3
Ø must be in II or III
Ø = 180 - 70.529 or Ø = 180 + 70.529
= 109.471° or 250.053°
I assume you know how to change those decimals to degrees and minutes.