Question

5cos²x-6cosx+1=0И 2ctgx-3tgx+1=0

Answer 1

5cos²x - 6cosx + 1 = 0,

cosx = а,

5а² - 6а + 1 = 0,

Д = (-6)² - 4*5*1 = 36 - 20 = 16,

а1 = (6 + 4) / 2*5 = 10/10 = 1,

а2 = (6 - 4) / 2*5 = 2/10 = 1/5,

cosx = а1,

cosx = 1,

х1 = 2πn, n ∈ Z,

cosx = а2,

cosx = 1/5,

х2 = ±arccos (1/5) + 2πn, n ∈ Z,

2ctgx - 3tgx + 1 = 0,

2/(tgx) - 3tgx + 1 = 0,   (* tgx)

2tgx - 3tg²x + 1 = 0,

3tg²x - 2tgx - 1 = 0,

tgx = а,

3а² - 2а - 1 = 0,

Д = (-2)² - 4*3*(-1) = 4 + 12 = 16,

а1 = (2 + 4) / 2*3 = 6/6 = 1,

а2 = (2 - 4) / 2*3 = -2/6 = -1/3,

tgx = а1,

tgx = 1,

х = arctg1 + πn, n ∈ Z,

x = π/4 + πn, n ∈ Z,

tgx = а2,

tgx = -1/3,

∅

Answer 2

To solve the equations:

1. 5cos²x - 6cosx + 1 = 0

We can rewrite this equation as:

5(cos²x - (6/5)cosx) + 1 = 0

Now, let's make a substitution. Let u = cosx. The equation becomes:

5u² - (6/5)u + 1 = 0

Multiplying the equation by 5 to eliminate the denominator, we get:

25u² - 6u + 5 = 0

Now we can solve this quadratic equation using the quadratic formula:

u = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 25, b = -6, and c = 5. Substituting these values, we get:

u = (-(-6) ± √((-6)² - 4 * 25 * 5)) / (2 * 25) = (6 ± √(36 - 500)) / 50 = (6 ± √(-464)) / 50

Since the discriminant (b² - 4ac) is negative, there are no real solutions for u. Therefore, there are no real solutions for the equation 5cos²x - 6cosx + 1 = 0.

2. 2ctgx - 3tgx + 1 = 0

We can rewrite this equation as:

2(cotx - (3/2)tanx) + 1 = 0

Now, let's make a substitution. Let u = tanx. The equation becomes:

2(cotx - (3/2)u) + 1 = 0

Multiplying through by 2 to eliminate the fraction, we get:

cotx - 3u + 2 = 0

Rearranging the equation, we have:

cotx = 3u - 2

Taking the reciprocal of both sides:

tanx = 1 / (3u - 2)

Now we have a relation between tanx and u. However, it doesn't provide a direct solution for x.

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