2 root 2 sin 2t + 2 root 3 tan 2t = 0
To solve the equation 2√2sin(2t) + 2√3tan(2t) = 0, we'll need to simplify and manipulate the equation to express it in terms of a single trigonometric function. Then we can solve for t.
Let's begin by simplifying the equation term by term.
First, we can rewrite the expression 2√2sin(2t) as √8sin(2t) by factoring out the square root of 2 from the coefficient.
Now the equation becomes √8sin(2t) + 2√3tan(2t) = 0.
Next, we know that tan(2t) is equal to sin(2t)/cos(2t), so we can rewrite the equation as:
√8sin(2t) + 2√3(sin(2t)/cos(2t)) = 0.
To eliminate the fraction, we can multiply both sides of the equation by cos(2t):
√8sin(2t)cos(2t) + 2√3sin(2t) = 0.
Using the double angle identity sin(2θ) = 2sin(θ)cos(θ), we can simplify further:
√8(2sin(t)cos(t))cos(2t) + 2√3sin(2t) = 0.
Simplifying:
4√2sin(t)cos(t)cos(2t) + 2√3sin(2t) = 0.
We can notice that sin(2t) can be written as 2sin(t)cos(t) by another double angle identity:
4√2(2sin(t)cos(t))cos(2t) + 2√3sin(2t) = 0.
This further simplifies to:
8√2sin(t)cos(t)cos(2t) + 2√3(2sin(t)cos(t)) = 0.
Factoring out a common term of 2√2sin(t)cos(t), we get:
2√2sin(t)cos(t)(4cos(2t) + √3) = 0.
Now we can solve this equation by setting each factor equal to zero:
2√2sin(t)cos(t) = 0 or 4cos(2t) + √3 = 0.
First, let's solve 2√2sin(t)cos(t) = 0:
This equation is satisfied when sin(t) = 0 or cos(t) = 0.
If sin(t) = 0, then t can be any integer multiple of π.
If cos(t) = 0, then t can be (n + 1/2)π, where n is an integer.
Now let's solve 4cos(2t) + √3 = 0:
Subtracting √3 from both sides, we have:
4cos(2t) = -√3.
Dividing both sides by 4, we get:
cos(2t) = -√3/4.
The values of t that satisfy this equation can be found using the inverse cosine function (arccos):
2t = arccos(-√3/4).
Solving for t:
t = (1/2) * arccos(-√3/4) + kπ
where k is an integer.
So, the solutions to the equation 2√2sin(2t) + 2√3tan(2t) = 0 are:
1. t = nπ, where n is an integer.
2. t = (1/2) * arccos(-√3/4) + kπ, where k is an integer.
These are the general solutions for the equation.