Find all solutions to the equation tan(t)=1/tan (t) in the interval 0<t<2pi.

Solve the equation in the interval [0, 2pi]. The answer must be a multiple of pi
2sin(t)cos(t) + sin(t) -2cos(t)-1=0

Find all solutions of the equation 2cos3x=1

tan(t)=1/tan (t)

tan^2 t = 1
tan t = ± 1
t = π/4, 3π/4, 5π/4, and 7π/4

2sin(t)cos(t) + sin(t) -2cos(t)-1= 0
by grouping:
sint(2cost + 1) - (2cost + 1) = 0
(2cost+1)(sint-1) = 0
cost = -1/2 or sint = 1

I will let you finish that one

2cos3x = 1
cos3x = 1/2
3x = π/3 or 3x = 2π-π/3 = 5π/3

period of cos3x = 2π/3

so x = π/3 + 2kπ/3
x = 2π/3 + 2kπ/3 , where k is an integer

thanks but it helped some but not much. Reiny

To find all solutions to the equation tan(t) = 1/tan(t) in the interval 0 < t < 2pi, we can first rewrite the equation as:

tan^2(t) = 1

Then, using the identity tan^2(t) + 1 = sec^2(t), we have:

sec^2(t) = 1

Taking the square root of both sides, we get:

sec(t) = ±1

Since the secant function is the reciprocal of the cosine function, this equation can be rewritten as:

1/cos(t) = ±1

Simplifying further, we have:

cos(t) = ±1

The cosine function takes values of 1 and -1 at multiples of pi, so the solutions to the equation are:

t = π/2 + kπ, where k is an integer

However, since the given interval is 0 < t < 2pi, we need to filter out any solutions that fall outside this range. Thus, the solutions that satisfy the given conditions are:

t = π/2, 3π/2

To find all the solutions to the equation tan(t) = 1/tan(t) in the interval 0 < t < 2pi, we can use the properties of the tangent function.

First, let's solve for tan(t):

tan(t) = 1/tan(t)

Multiply both sides by (tan(t))^2 to get rid of the fraction:

(tan(t))^2 = 1

Now, take the square root of both sides:

tan(t) = ±1

Since we want solutions in the interval 0 < t < 2pi, we can use the unit circle and the quadrant signs of the tangent function.

In the first quadrant (0 < t < pi/2), tan(t) > 0, so we can ignore the negative sign. The positive solution is tan(t) = 1, which gives us t = pi/4.

In the second quadrant (pi/2 < t < pi), tan(t) < 0, so again, we can ignore the negative sign. The positive solution is tan(t) = -1, which gives us t = 3pi/4.

In the third quadrant (pi < t < 3pi/2), tan(t) > 0, so we ignore the negative sign. The positive solution is tan(t) = 1, which gives us t = 5pi/4.

In the fourth quadrant (3pi/2 < t < 2pi), tan(t) < 0, so we ignore the negative sign. The positive solution is tan(t) = -1, which gives us t = 7pi/4.

Therefore, the solutions in the interval 0 < t < 2pi are t = pi/4, 3pi/4, 5pi/4, and 7pi/4.

Now, let's solve the equation 2sin(t)cos(t) + sin(t) - 2cos(t) - 1 = 0.

First, let's rewrite the equation:

sin(t)(2cos(t) + 1) - cos(t)(2cos(t) + 1) = 0

Factor out the common factor:

(2cos(t) + 1)(sin(t) - cos(t)) = 0

Now set each factor equal to zero:

1) 2cos(t) + 1 = 0
2cos(t) = -1
cos(t) = -1/2

Taking the inverse cosine of both sides, we get:

t = π + π/3 = 4π/3

2) sin(t) - cos(t) = 0
sin(t) = cos(t)

Divide both sides by cos(t):

tan(t) = 1

Using the equation from the previous question, we know the solutions to tan(t) = 1 are t = π/4 and 5π/4.

Therefore, the solutions to the equation 2sin(t)cos(t) + sin(t) - 2cos(t) - 1 = 0 in the interval [0, 2pi] are t = 4π/3, π/4, and 5π/4.

Finally, let's solve the equation 2cos(3x) = 1:

First, divide both sides by 2:

cos(3x) = 1/2

We want the solutions in the interval [0, 2pi], so we can use the unit circle to find the angles where cosine is equal to 1/2.

For cosine, we have the following solutions in the interval [0, 2pi]:

3x = π/3 and 3x = 5π/3

Solving for x, we get:

x = π/9 and x = 5π/9

Therefore, the solutions to the equation 2cos(3x) = 1 in the interval [0, 2pi] are x = π/9 and 5π/9.