2sinx * sin2x - sinx =0

Предыдущий
вопрос Следующий
вопрос

To solve the equation 2sin(x) * sin(2x) - sin(x) = 0, we can simplify it by using trigonometric identities.

First, rewrite sin(2x) in terms of sin(x) using the double angle identity:

sin(2x) = 2sin(x)cos(x)

Now, substitute sin(2x) = 2sin(x)cos(x) back into the original equation:

2sin(x) * 2sin(x)cos(x) - sin(x) = 0
4sin^2(x)cos(x) - sin(x) = 0

Factor out sin(x) from the equation:

sin(x)(4sin(x)cos(x) - 1) = 0

Now we have two possible solutions:

1) sin(x) = 0
This means that x can be any multiple of pi since sin(pi) = 0.

2) 4sin(x)cos(x) - 1 = 0
Using the trigonometric identity for sin(2x) again:
2sin(x)cos(x) = 1

sin(2x) = 1
2x = pi/2
x = pi/4

Therefore, the solutions to the equation 2sin(x) * sin(2x) - sin(x) = 0 are x = pi/4 and x = npi where n is an integer.