Question

Sin^x-5sin x*cos x+4cos^ x=0

Answer 1

Если правильно понимаю,что sin^x,cos^x-имеешь ввиду в квадрате,то
делим обе части на соs^x,
tg^x-5tg x+4=0
1)пусть tg x-это а,тогда
а^2-5а+4=0
D=b^2-4ac=25-16=9
a1=-b+√D/2a=5+3/2=4
a2=-b-√D/2a=5-3/2=1
2)tg x=1 или tgx=4
x=π/4. x-не сущест.

Answer 2

We can simplify the equation using trigonometric identities. Let's start by factoring out sin(x) from the first two terms and cos(x) from the last term:

sin(x) - 5sin(x)cos(x) + 4cos^2(x)sin(x) = 0

Now we can use the identity cos^2(x) = 1 - sin^2(x) to replace cos^2(x) with 1 - sin^2(x):

sin(x) - 5sin(x)cos(x) + 4(1 - sin^2(x))sin(x) = 0

Expanding the last term:

sin(x) - 5sin(x)cos(x) + 4sin(x) - 4sin^3(x) = 0

Collecting like terms:

sin(x)(1 - 5cos(x) + 4 - 4sin^2(x)) = 0

We can see that sin(x) = 0 is one solution. For the other solution, we need to solve the quadratic expression in the parentheses:

1 - 5cos(x) + 4 - 4sin^2(x) = 0

Simplifying:

-4sin^2(x) - 5cos(x) + 5 = 0

Using the identity sin^2(x) + cos^2(x) = 1, we can replace sin^2(x) with 1 - cos^2(x):

-4(1 - cos^2(x)) - 5cos(x) + 5 = 0

Expanding and rearranging:

4cos^2(x) + 5cos(x) - 1 = 0

Solving this quadratic equation using the quadratic formula, we get:

cos(x) = (-5 ± sqrt(41))/8

Therefore, the solutions to the original equation are:

sin(x) = 0 or cos(x) = (-5 ± sqrt(41))/8