Question

Sin2x+1=0

Answer 1

sin2x +1 =0

sin2x = -1

2x = -пи\2 +2пи*n( частный случай)

x = -пи\4 + пи*n

Answer 2

To solve the equation sin(2x) + 1 = 0, we can follow these steps:

1. Subtract 1 from both sides of the equation: sin(2x) = -1

2. Now, we need to find the angles where the sine function equals -1. In the unit circle, the sine function is -1 at two angles: -π/2 and 3π/2.

3. We can write the equation as: 2x = -π/2 + 2πn (where n is an integer) or 2x = 3π/2 + 2πn (where n is an integer)

4. Solve for x in each equation: For the first equation: 2x = -π/2 + 2πn x = (-π/2 + 2πn)/2 x = -π/4 + πn (where n is an integer)

   For the second equation: 2x = 3π/2 + 2πn x = (3π/2 + 2πn)/2 x = 3π/4 + πn (where n is an integer)

So, the solutions to the equation sin(2x) + 1 = 0 are: x = -π/4 + πn (where n is an integer) x = 3π/4 + πn (where n is an integer)