Question

1+sinx·cosx-3cos²x=0 6cos²x-2sin2x=1 tg²x-3tgx+4=3ctgx-ctg²x sinx·cosx-cos²x=1

Answer 1

1+sinx·cosx-3cos²x=0
sin^2x+cos^2x+sinxcosx-3cos^2x=0/cos^2x
tg^x+1+tgx-3=0
a^2+a-2=0
D=3
a1=-2
a2=-1 x=-pi/4+pin 3) sinxcosx-cos^2x=1 tgx-1=1 tgx=2 x=arctg2+pin

Answer 2

It looks like you have written down several trigonometric equations. I can help you solve them one by one.

1. 1+sinx·cosx-3cos²x=0

We can simplify this equation using the identity sin^2x + cos^2x = 1.

sinx·cosx - 3cos²x + 1 = sinx·cosx - (1+sin^2x) + 1 sinx·cosx - sin^2x = 0 sinx·(cosx - sinx) = 0

So either sinx = 0 or cosx = sinx. If sinx = 0, then x = nπ (where n is an integer) is a solution. If cosx = sinx, then x = π/4 + nπ/2 (where n is an integer) is a solution.

2. 6cos²x - 2sin2x = 1

We can simplify sin2x using the identity sin2x = 2sinx·cosx.

6cos²x - 4sinx·cosx + 1 = 0 (2cosx - 1)(3cosx + 1) = 0

So either cosx = 1/2 or cosx = -1/3. If cosx = 1/2, then x = π/3 + 2nπ or x = 5π/3 + 2nπ (where n is an integer) are solutions. If cosx = -1/3, then x = arccos(-1/3) + 2nπ or x = -arccos(-1/3) + (2n+1)π (where n is an integer) are solutions.

3. tg²x-3tgx+4=3ctgx-ctg²x

Let's start by using the identity 1 + tan^2x = sec^2x and 1 + cot^2x = cosec^2x.

sec^2x - 3tanx + cosec^2x = 0 (1 + cos^2x)/cos^2x - 3sinx/cosx + (1 + sin^2x)/sin^2x = 0 (1 + cos^2x)/cos^2x - 3sinx/cosx + (1 + sin^2x)/sin^2x = 0

Multiplying through by sin^2x·cos^2x, we get:

sin^2x + cos^2x - 3sinx·cosx + sin^2x + cos^2x = 0 2sin^2x - 3sinx·cosx = 0 sinx·(2sinx - 3cosx) = 0

So either sinx = 0 or 2sinx = 3cosx. If sinx = 0, then x = nπ (where n is an integer) is a solution. If 2sinx = 3cosx, then x = arctan(2/3) + nπ or x = π - arctan(2/3) + nπ (where n is an integer) are solutions.

4. sinx·cosx - cos²x = 1

We can simplify this equation using the identity sin^2x + cos^2x = 1.

sinx·cosx - (1-sin^2x) = 1 sinx·cosx +