What are the solutions to the equation 3cos(t)+4=2 on the interval [0, pi]?

Question

Find all solutions to 3cos(t)+4 = 2 on the interval [0, π) without the use of a graphing utility. If there is no solution, write N/A. Solve the question algebraically, listing each step in turn.

Answer 1

3cost + 4 = 2

3cost +4-4 = 2-4 = -2

cost = -2/3

t = arccos(-2/3) which is an angle in quadrant 2 or 3 where all cosines are negative

Answer 2

Let's solve the equation 3cos(t) + 4 = 2 for t in the interval [0, π]:

1. First, we isolate the cosine term. We subtract 4 from both sides of the equation:

3cos(t) = 2 - 4

Simplifying, we get:

3cos(t) = -2

2. Next, we solve for cos(t) by dividing both sides of the equation by 3:

cos(t) = -2/3

3. We now need to find the angle t for which cos(t) = -2/3. We can do this using the inverse cosine function, often written as cos^(-1) or arccos:

t = cos^(-1)(-2/3)

The cosine function gives values in the range [-1, 1], and -2/3 is within this range. In the interval [0, π], the cosine function can take any value from -1 to 1. Therefore, there is a solution in this interval.

The exact solution to the equation is t = cos^(-1)(-2/3), which is approximately 2.3 radians. This is the angle in the interval [0, π] for which the cosine is -2/3.

Answer 3

3cos(t)+4=2

cos(t)=-2/3 Puts t in quadrant 2 on [0,π]

cos^-1(-2/3) =131.8°=2.3 rad=0.7322π rad