Question

Sin^2(pi+x)-cos(-x-3pi/2)=0​

Answer 1

Ответ: x₁=πn     x₂=π/2+2πn.

Объяснение:

sin^2(pi+x)-cos(-x-3pi/2)=0​

(-sinx)²-cos(-(x+3π/2))=0

sin²x-cos(3π/2+x)=0

sin²x-sinx=0

sinx*(sinx-1)=0

sinx=0

x₁=πn.

sinx-1=0

sinx=1

x₂=π/2+2πn.

Answer 2

Let's simplify the expression using trigonometric identities:

sin^2(pi+x) - cos(-x-3pi/2) = 0

First, let's simplify sin^2(pi+x) using the identity:

sin^2(theta) + cos^2(theta) = 1

sin^2(pi+x) = 1 - cos^2(pi+x)

sin^2(pi+x) = 1 - cos(pi+x)^2

Now let's simplify cos(-x-3pi/2) using the identity:

cos(theta + pi) = -cos(theta)

cos(-x-3pi/2) = -cos(x+pi/2)

Substituting these simplifications into the original expression, we get:

1 - cos(pi+x)^2 + cos(x+pi/2) = 0

Now let's use the identity:

cos(theta + pi/2) = -sin(theta)

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

Substituting this into the expression, we get:

1 - cos(pi+x)^2 - sin(x) = 0

Now let's use the identity:

cos(theta)^2 + sin(theta)^2 = 1

cos(pi+x)^2 + sin(x)^2 = 1

Substituting this into the expression, we get:

1 - 1 - sin(x) = 0

-sin(x) = 0

sin(x) = 0

Therefore, the solutions for x are:

x = n*pi, where n is an integer.