f(x) = tan2(x) + (√3 - 1)[tan(x)] - √3 = 0
tan2(x) + √3[tan(x)] - tan(x) - √3 = 0
Factor into
[-1 + tan(x)]*[√3 + tan(x)] = 0
which means
[-1 + tan(x)] = 0 and/or [√3 + tan(x)] = 0
Then
tan(x) = 1
tan-1(1) = pi/4 radians
For the other equation
[√3 + tan(x)] = 0
tan(x) = -√3
tan-1(-√3) = -pi/3
so that
x = pi/4 or -pi/3 in the interval [0, 2pi]
Answer:
yea
Step-by-step explanation:
Answer:
hi
Step-by-step explanation:
Answer:
j = -6
Step-by-step explanation:
j + 2j + 3j = -20 + (-16)
6j = -36
j = -6
Ok me pls thank you so much you are a heavenly angel
