Question

Sin^2x+sinxsin(3п/2-x)-cos2x=1

Answer 1

Sin^2x+sinxsin(3п/2-x)-cos2x=1
sin²x-sinxcosx-cos²x+sin²x-sin²x-cos²x=0
sin²x-sinxcosx-2cos²x=0/cos²x
tg²x-tgx-2=0
tgx=a
a²-a-2=0
a1+a2=1 U a1*a2=-2
a1=-1⇒tgx=-1⇒x=-π/4+πk,k∈z
a2=2⇒tgx=2⇒x=arctg2+πk,k∈z

Answer 2

Cos^2x-sinxcosx=sin^2x+cos^2x
sin^2x+sinxcosx=0
sinx(sinx+cosx)=0
sinx=0
x=Пk k=-3 x=-3П
k=-2 x=-2П
sinx=-cosx
tgx=-1
x=-П/4+Пk
k=-2
x=-9П/4

Answer 3

Let's simplify the left-hand side of the equation:

sin^2x + sinxsin(3π/2 - x) - cos2x

Using the identity sin(π/2 - x) = cos(x), we can rewrite sin(3π/2 - x) as cos(x):

sin^2x + sinxcosx - cos2x

Using the identity cos2x = cos^2x - sin^2x, we can rewrite cos2x as:

cos^2x - sin^2x - sinxcosx

Now we have:

sin^2x + sinxcosx - cos^2x + sin^2x + sinxcosx = 1

Using the identity sin^2x + cos^2x = 1, we can simplify further:

2sin^2x + 2sinxcosx = 1

Using the identity 2sinxcosx = sin2x, we get:

2sin^2x + sin2x = 1

Now we can rearrange the terms:

2sin^2x + sin2x - 1 = 0

This is a quadratic equation in sinx. We can solve for sinx using the quadratic formula:

sinx = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 2, b = 1, and c = -1. Substituting these values, we get:

sinx = (-1 ± sqrt(1 + 8)) / 4

sinx = (-1 ± sqrt(9)) / 4

We have two solutions:

sinx = 1 or sinx = -1/2

To check these solutions, we can substitute them back into the original equation:

For sinx = 1:

sin^2x + sinxsin(3π/2 - x) - cos2x = 1

1 + sin(3π/2 - x) - cos2x = 1

Using the identity sin(π/2 - x) = cos(x), we get:

1 + cosx - cos2x = 1

Using the identity cos2x = cos^2x - sin^2x, we get:

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

Simplifying, we get:

sin^2x + cosx = 0

This is not true for sinx = 1, so this solution is not valid.

For sinx = -1/2:

sin^2x + sinxsin(3π/2 - x) - cos2x = 1

(1/4) - (1/2)cosx - cos2x = 1

Using the identity cos2x = cos^2x - sin^2x, we get:

(1/4) - (1/2)cosx - (cos^2x - 1 + cos^2x) = 1

Simplifying, we get:

2cos^2x + cosx - 5/4 = 0

Using the quadratic formula again, we get:

cosx = (-1 ± sqrt(1 + 40)) / 4

cosx = (-1 ± sqrt(41)) / 4

We have two solutions:

cosx ≈ 0.882 or cosx ≈ -1.132

Substituting these values back into the original equation, we can verify that cosx ≈ 0.882 is a valid solution, but cosx ≈ -1.132 is not.

Therefore, the only valid solution is