1) sin2x = sin2x
2) tan2x +tanx = 0
vicky
answered
#=alpha
Sin:
x=#+2kpi and x=pi-#+2kpi
When u come out with the value of sine (alpha) you place it in the both cases and then you put it in the interval
0 <#+2kpi <2pi
Steve
answered
#2
tan2x + tanx = 0
2tanx/(1-tan^2(x)) + tanx = 0
2tanx + tanx(1-tan^2(x)) = 0
tanx (2 + 1-tan^2x) = 0
tanx=0
tanx = ±√3
so, x=kπ or kπ ± π/3
Now just use the values of k so that the results come out in the desired domain.
Explain Bot
answered
1) sin2x = sin2x:
This equation is an identity, which means it is always true for any value of x. In other words, it holds for all x in the given interval [0,2π). So, every value of x in the interval is a solution for this equation. Therefore, the solution set for this equation is the entire interval [0,2π).
2) tan2x + tanx = 0:
To find solutions for this equation, let's simplify it first.
Using the trigonometric identity tan2x = (2tanx) / (1-tan^2x), we can rewrite the equation as:
(2tanx) / (1-tan^2x) + tanx = 0.
Next, let's find a common denominator and combine the terms:
[(2tanx) + (tan^2x)(1-tan^2x)] / (1-tan^2x) = 0
Simplifying further:
(2tanx + tan^2x - tan^4x) / (1 - tan^2x) = 0.
Now, we need to identify the values that make the numerator or the denominator equal to zero.
For the numerator:
2tanx + tan^2x - tan^4x = 0
This equation can be factored as follows:
tanx(tanx+1)(tan^2x-1) = 0.
This gives us three separate equations to solve:
tanx = 0, tanx+1 = 0, tan^2x - 1 = 0.
Solving each equation, we get the solutions:
tanx = 0:
x = 0, π.
tanx + 1 = 0:
tanx = -1,
x = -π/4, 3π/4.
tan^2x - 1 = 0:
tanx = ±1,
x = π/4, 5π/4.
For the denominator:
1 - tan^2x = 0.
This equation can be rearranged as:
tan^2x = 1.
Taking the square root of both sides, we have:
tanx = ±1,
which gives us two additional solutions:
x = π/4, 3π/4.
To summarize, the solutions for the equation tan2x + tanx = 0 in the interval [0,2π) are:
x = 0, π, -π/4, 3π/4, π/4, 5π/4.
I hope this helps! Let me know if you have any further questions.
