2 root 2 sin 2t + 2 root 3 tan 2t = 0
To solve the equation 2√2sin(2t) + 2√3tan(2t) = 0, we need to find the values of t that satisfy this equation.
First, let's simplify the equation by factoring out the common term of 2:
2(√2sin(2t) + √3tan(2t)) = 0
Now, we have two cases to consider:
Case 1: √2sin(2t) + √3tan(2t) = 0
We can start by factoring out the common term of √2:
√2(sin(2t) + √3/√2 * tan(2t)) = 0
Now, we have two possibilities:
1. √2 = 0 (which is not possible)
2. sin(2t) + √3/√2 * tan(2t) = 0
To solve this equation, we need to express tan(2t) in terms of sin(2t). We know that tan(2t) = sin(2t) / cos(2t). Since √3/√2 = sin(π/3), we can rewrite the equation as:
sin(2t) + sin(π/3) * sin(2t) / cos(2t) = 0
Next, we can multiply both sides by cos(2t) to eliminate the denominator:
sin(2t)cos(2t) + sin(π/3) * sin(2t) = 0
Using the double angle identity for sin(2t): sin(2t) = 2sin(t)cos(t), we can substitute and simplify the equation:
2sin(t)cos(t)cos(2t) + sin(π/3) * 2sin(t)cos(t) = 0
Factoring out 2sin(t)cos(t):
2sin(t)cos(t)(cos(2t) + sin(π/3)) = 0
Now we have two possibilities:
1. sin(t) = 0 (which means t = 0, π, 2π, ...)
2. cos(2t) + sin(π/3) = 0
To solve the equation cos(2t) + sin(π/3) = 0, we can rearrange it as cos(2t) = -sin(π/3). We know that sin(π/3) = √3/2, so we have:
cos(2t) = -√3/2
Now, find the values of 2t that satisfy this equation by taking the inverse cosine (arccos) of both sides:
2t = arccos(-√3/2)
To get the general solutions, we need to consider the unit circle and the periodic nature of cosine function. The principal range of the arccos function is [0, π], so we have:
1. 2t = arccos(-√3/2) → t = arccos(-√3/2) / 2
Since cosine function is negative in the second and third quadrants, we can add π to the angle to obtain more solutions:
2. t = (arccos(-√3/2) + π) / 2
To summarize, the solutions for the given equation are:
t = arccos(-√3/2) / 2 + 2πk (k is an integer)
t = (arccos(-√3/2) + π) / 2 + 2πk (k is an integer)