Question

1)cosx-sinx=4 cosx sin^2x;2)(корень из 3) - 0.5sin2x=0помогите пожалуйста

Answer 1

1)cosx-sinx=2sin2xsinx
cosx-sinx=2*1/2(cosx-cos3x)
cosx-sinx-cosx+cos3x=0
cos3x-sinx=0
cos3x-cos(π/2-x)=0
-2sin(2x-π/4)sin(x+π/4)=0
sin(2x-π/4)=0⇒2x-π/4=πn⇒2x=π/4+πn⇒x=π/8+πn/2
sin(x+π/4)=0⇒x+π/4=πn⇒x=-π/4+πn
2)1/2sin2x=√3⇒sin2x=2√3∉[-1;1]-нет решения

Answer 2

1. To solve the equation cos(x) - sin(x) = 4cos(x)sin^2(x), we can use trigonometric identities to simplify the left-hand side of the equation:

cos(x) - sin(x) = cos(x)cos(π/2) - sin(x)sin(π/2) (using the identity cos(a - b) = cos(a)cos(b) + sin(a)sin(b) with a = x and b = π/2) = cos(x + π/2)

Then, substituting this expression into the original equation, we get:

cos(x + π/2) = 4cos(x)sin^2(x)

Using the double angle formula for sine, sin(2x) = 2sin(x)cos(x), we can rewrite the right-hand side of the equation as:

4cos(x)sin^2(x) = 2(2cos(x)sin(x))sin(x) = 2sin(2x)sin(x)

Substituting this expression into the equation, we get:

cos(x + π/2) = 2sin(2x)sin(x)

Using the sum-to-product formula for cosine, cos(a + b) = cos(a)cos(b) - sin(a)sin(b) with a = x and b = π/2, we can rewrite the left-hand side of the equation as:

cos(x + π/2) = cos(x)cos(π/2) - sin(x)sin(π/2) = -sin(x)

Substituting this expression into the equation, we get:

-sin(x) = 2sin(2x)sin(x)

Dividing both sides by sin(x) (which is nonzero since sin(x) = 0 would imply cos(x) = 1, which is not consistent with the original equation), we get:

-1 = 2sin(2x)

Solving for sin(2x), we get:

sin(2x) = -1/2

Using the double angle formula for sine again, we get:

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

Using the identity sin^2(x) + cos^2(x) = 1, we can solve for cos(x):

cos^2(x) = 1 - sin^2(x) = 1 - (sin(x)cos(x))^2 = 1 - 1/16 = 15/16

Taking the square root of both sides (and choosing the positive square root since cos(x) is positive when sin(2x) = -1/2), we get:

cos(x) = sqrt(15)/4

Therefore, the solutions to the equation cos(x) - sin(x) = 4cos(x)sin^2(x) are given by:

x = π/4 ± kπ + arcsin(sqrt(15)/4), where k is an integer.

2. To solve the equation sqrt(3) - 0.5sin(2x) = 0, we can isolate sin(2x) on one side of the equation:

sin(2x) = 2sqrt(3)/5

Using the double angle formula for sine, we get:

2sin(x)cos(x) = 2sqrt(3)/5

Dividing both sides by 2cos(x) (which is nonzero since cos(x) = 0 would imply sin(2x) = 0, which is not consistent with the original equation), we get:

sin(x) = sqrt(3)/5cos(x)

Using the identity sin^2(x) + cos^2(x) = 1, we can solve