Question

Решите, пожалуйста sin2x cosx = cos2x sinx cos5x cosx = cos4x 3+sin2x = 4sin^2x cos2x + cos^2x + sinx cos x = 0 3 cos 2x + sin^2x + 5 sinx cosx = 0

Answer 1

sin2x cosx = cos2x sinx

2sinxcosxcosx=cos2xsinx    sinx=0   x=Пk

2cos^2x=cos2x

2cos^2x=2cos^2x-1   ∅

ответ x=Пk

cos5x cosx = cos4x

cos4x+cos6x=2cos4x

cos6x-cos4x=0

-2sin5xsinx=0

x=Пk

x=Пk/5

3+sin2x = 4sin^2x

3sin^2x+3cos^2x+2sinxcosx=4sin^2x

sin^2x-3cos^2x-2sinxcosx=0

sinx/cosx-3cosx/sinx-2=0

tgx-3/tgx-2=0

tg^2x-2tgx-3=0    tgx=3   tgx=-1

x=-П/4+Пk

x=arctg3+Пk

cos2x + cos^2x + sinx cos x = 0

2cos^2x-sin^2x+sinxcosx=0  |sinxcosx

2cosx/sinx-sinx/cosx+1=0

2ctgx-tgx+1=0

2/tgx-tgx+1=0

-tg^2x+tgx+2=0  tg^2x-tgx-2=0

tgx=(1+-3)/2 tgx=2  tgx=-1

x=-П/4+Пk

x=arctg2+Пk

3 cos 2x + sin^2x + 5 sinx cosx = 0

3cos^2x-2sin^2x+5sinxcosx=0

3cosx/sinx-2sinx/cosx+5=0

3/tgx-2tgx+5=0

2tgx-3/tgx-5=0

2tg^2x-5tgx-3=0

tgx=(5+-7)/4   tgx=3    tgx=-1/2

x=arctg3+Пk

x=-arctg1/2+Пk

Answer 2

1. sin2x cosx = cos2x sinx

Starting with the left-hand side:

sin2x cosx = 2sinxcosx * cosx (using the double angle formula for sine) = 2cosxsinxcosx = cosx(2sinxcosx)

Now, using the double angle formula for cosine:

cos2x sinx = (cos^2 x - sin^2 x)sinx = cosx(cosxsinx - sinxcosx) = cosx( sinx cosx - sinx cosx) = 0

Therefore, sin2x cosx = cos2x sinx is true for any value of x.

2. cos5x cosx = cos4x

Using the identity cos(a+b) = cos(a)cos(b) - sin(a)sin(b), we can write:

cos5x cosx = (cos4x cosx - sin4x sinx)cosx (applying the formula twice) = cos4x cosx^2 - sin4x sinx cosx = cos4x (1 - sin^2 x) - sin4x sinx

Now, using the identity sin^2 x + cos^2 x = 1 and simplifying:

cos5x cosx = cos4x - cos4x sin^2 x - sin4x sinx = cos4x(1-sin^2x) - sin4x sinx = cos4x cos^2 x - sin4x sinx = cos4x (1 - sin^2 x) - cos4x sin^2 x (applying the identity again) = cos4x

Therefore, cos5x cosx = cos4x is true for any value of x.

3. 3+sin2x = 4sin^2x cos2x + cos^2x + sinx cos x = 0

We can rearrange the equation as:

4sin^2x cos2x + cos^2x + sinx cos x = -3 - sin^2 x

Now, using the identity sin^2 x + cos^2 x = 1, we can write:

4sin^2x cos2x + cos^2x + sinx cos x = -(sin^2 x + 3 cos^2 x)

Using the identity cos2x = 1 - 2sin^2x, we can write:

4sin^2x (1 - 2sin^2x) + cos^2x + sinx cos x = -(sin^2 x + 3(1 - sin^2 x)) 4sin^2x - 8sin^4x + cos^2x + sinx cos x = -2sin^2 x 8sin^4x - 3sin^2x + cos^2x + sinx cos x + 2 = 0

Unfortunately, this equation cannot be easily solved algebraically, and we need to resort to numerical or graphical methods to find approximate solutions.

4. 3 cos 2x + sin^2x + 5 sinx cosx = 0

Using the identity cos2x = 1 - 2sin^2x, we can write:

3(1 - 2sin^2x) + sin^2x + 5sinx cosx = 0 3 - 6sin^2x + sin^2x + 5sinx cosx = 0 -5sin^