To solve this equation, we will first apply the trigonometric identity:
tg(-θ) = -tg(θ)
So we have:
√3tg(3x/4 - π/6) = -1
tg(3x/4 - π/6) = -1/√3
Now we need to find the angle whose tangent is -1/√3. We know that tg(π/6) = √3, so tg(-π/6) = -√3.
Therefore, the reference angle for -1/√3 is -π/6. However, we need to find the actual angle in the interval [0, 2π] that has this tangent value.
Since the value is negative, we are looking for the angle in the third quadrant where tangent is negative.
Thus, the angle is:
3x/4 - π/6 = -π/6
3x/4 = 0
x = 0
So, the solution to the equation is x = 0.
To solve this equation, we will first apply the trigonometric identity:
tg(-θ) = -tg(θ)
So we have:
√3tg(3x/4 - π/6) = -1
tg(3x/4 - π/6) = -1/√3
Now we need to find the angle whose tangent is -1/√3. We know that tg(π/6) = √3, so tg(-π/6) = -√3.
Therefore, the reference angle for -1/√3 is -π/6. However, we need to find the actual angle in the interval [0, 2π] that has this tangent value.
Since the value is negative, we are looking for the angle in the third quadrant where tangent is negative.
Thus, the angle is:
3x/4 - π/6 = -π/6
3x/4 = 0
x = 0
So, the solution to the equation is x = 0.