Question

1)2sinx-1=02)sin(4x-π/4)=0

Answer 1

1) 2sinx=1
Sinx=1/2
X=(-1)^n*arcsin1/2+Пn
X=(-1)^n*П/6+Пn

2) (4x-П/4)=Пn
4x=П/4+Пn
X=П/16+Пn/4

Answer 2

1. To solve the equation 2sinx-1=0, we can isolate sinx by adding 1 to both sides, which gives:

2sinx = 1

Then, dividing both sides by 2:

sinx = 1/2

We know that the sine function is positive in the first and second quadrants, and that sin(30°) = 1/2. Therefore, the solutions for this equation are:

x = 30° + 360°k or x = 150° + 360°k, where k is an integer.

2. To solve the equation sin(4x-π/4)=0, we can use the fact that the sine function equals zero when the angle is a multiple of π:

4x - π/4 = nπ, where n is an integer.

Solving for x, we get:

x = (nπ + π/4)/4

We can simplify this expression by using the fact that π/4 = 4π/16:

x = (nπ + 4π/16)/4

x = (nπ + π/4)/4

So the solutions for this equation are:

x = (nπ + π/4)/4, where n is an integer.