Question

Sin2x + 2 sin x= 0 ??????????????????/

Answer 1

sin2X + 2sinX =0 

2sinX * cosX + 2sinX = 0 

2SinX(1+cosX)=0
sinX=0, то х=(число Пи) * k; 
сosx=-1, то х = (Число Пи) + (Число Пи) * n.

k и n - целые числа. 

Answer 2

To solve the equation sin(2x) + 2sin(x) = 0, we can use trigonometric identities and algebraic manipulation. Let's go through the steps:

Step 1: Use the double angle identity for sine: sin(2x) = 2sin(x)cos(x).

Step 2: Substitute sin(2x) with 2sin(x)cos(x) in the original equation: 2sin(x)cos(x) + 2sin(x) = 0.

Step 3: Factor out common terms: 2sin(x)(cos(x) + 1) = 0.

Step 4: Set each factor equal to zero and solve for x: a) 2sin(x) = 0 b) cos(x) + 1 = 0

For a): If 2sin(x) = 0, then sin(x) = 0. The solutions for this part are x = 0 and x = π (plus any integer multiple of 2π since sin(0) = sin(π) = 0).

For b): If cos(x) + 1 = 0, then cos(x) = -1. The solution for this part is x = π (plus any integer multiple of 2π since cos(π) = -1).

So, the complete set of solutions for the given equation is: x = 0 + 2nπ, where n is an integer x = π + 2nπ, where n is an integer

Note: This means that any integer multiple of 2π can be added to the solutions to account for all possible values of x that satisfy the equation.