Question

2 sin ^2 x cos^2 x - 5 sin x cos x =0

Answer 1

2 sin^2x + cos^2x - 5 sin x cos x = 0
2 sin^2x -( 1 - sin^2 x) - 5 sin x * cos x  = 0
3 sin^2 x - 5 sin x * cos x = 1
sin x ( 3 sin x - 5 cos x) = 1
sin x =1
x = p/2 + 2pm, m ∈ Z.

3 sin x - 5 cos x = 1
Тогда по формуле двойного угла выразим синус и косинус через тангенс:
Выразим тангенс через t: t = tg x/2,
ОДЗ: x/2 ≠ p/2 + pn, n ∈ Z.
x≠ p + 2pn, n ∈ Z.
1 + tg^2 x/2 ≠ 0
 
(3 t + 5 t^2 - 5)/ (1 + t^2) = 1
5t^2 - t^2 + 3t - 5-1 = 0
4t^2 + 3t - 6 = 0
D = 9+ 6*4*4 =105;
t(1,2) = -3 ±√105/ 8; 

(-3 - √105) / 8 = a   (-3 + √105) / 8 = b
tg x/2 = (-3 - √105) / 8;
x/2 = arctg a + pn, n ∈ Z.
x= 1/2 arctg a + 1/2 pn, n ∈ Z.

x/2 = arctg b + pk, k ∈ Z.
x= 1/2 arctg b + 1/2 pk, k ∈ Z.

Answer 2

To solve the equation 2sin^2(x)cos^2(x) - 5sin(x)cos(x) = 0, we can factor out common terms.

First, let's rewrite the equation using trigonometric identities:

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

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

2sin(x)cos(x)(sin(x)cos(x)) - 5sin(x)cos(x) = 0

Now, we can factor out the common term sin(x)cos(x):

sin(x)cos(x)(2sin(x)cos(x) - 5) = 0

Now, we have two possibilities:

1. sin(x)cos(x) = 0

This equation holds true if either sin(x) = 0 or cos(x) = 0. So we have two sets of solutions:

sin(x) = 0 => x = nπ, where n is an integer.

cos(x) = 0 => x = (2n + 1)π/2, where n is an integer.

2. 2sin(x)cos(x) - 5 = 0

To solve this equation, we can rewrite it as:

sin(2x) = 5/2

Now, we need to find the values of 2x for which the sine function equals 5/2. However, the sine function only takes values between -1 and 1, so there are no solutions for this equation.

Therefore, the solutions to the original equation 2sin^2(x)cos^2(x) - 5sin(x)cos(x) = 0 are:

x = nπ, where n is an integer.

x = (2n + 1)π/2, where n is an integer.