How do I solve 2sin4x cosx = 0 in the interval [0, 2pi)?
well, either
sin4x=0
or
cosx=0
so, 4x = 0,pi,2pi,3pi,4pi,...
meaning x = n*pi/4 for n=0 to 7
or, x = pi/2 or 3pi/2
but those are already covered by sin4x=0
see
http://www.wolframalpha.com/input/?i=+2sin4x+cosx
every time sin 4x = 0
and every time cos x = 0
sin 4 x = 0 when
x = 0, pi/4 , pi/2 , 3 pi/4, pi , 5 pi/4.... 8 pi/4 which is 2 pi
cos x = 0 when x = pi/2 , 3 pi/2 but we already have those two points
To solve the equation 2sin(4x)cos(x) = 0 in the interval [0, 2π), we need to find the values of x that satisfy this equation.
To solve this trigonometric equation, we can use the zero-product property, which states that if a product of factors equals zero, then at least one of the factors must be zero.
In this case, we have two factors: 2sin(4x) and cos(x). So, we can set each factor equal to zero and solve for x separately.
Factor 1: 2sin(4x) = 0
To find the values of x that satisfy this equation, we need to find when sin(4x) = 0. The sine function equals zero at x = 0, π, 2π, 3π, etc., and all their multiples.
Factor 2: cos(x) = 0
To find the values of x that satisfy this equation, we need to find when cos(x) = 0. The cosine function equals zero at x = π/2, 3π/2, etc.
Now, we need to find the common values of x that satisfy both factors. In this case, we are looking for the intersection points of the solutions of each factor.
From factor 1, sin(4x) = 0, we found x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, etc. These points might or might not be part of the interval [0, 2π), so we need to check.
From factor 2, cos(x) = 0, we found x = π/2, 3π/2. Now, we need to check if these points are within the interval [0, 2π).
Checking the values within the interval [0, 2π), we find that the common solutions for the equation 2sin(4x)cos(x) = 0 are:
- x = π/2
- x = 3π/2
Therefore, the equation 2sin(4x)cos(x) = 0 has two solutions in the interval [0, 2π): x = π/2 and x = 3π/2.