Question

Cos2x-cosx=0........................

Answer 1

Cos2x-cosx=0
2cos^2x-cosx-1=0
cosx=t
2t^2-t-1=0
t1=-0,5
t2=1
Делаем возврат,cosx=1,cosx=-0,5
x=2Пn,n принадлежит Z.
x=+-arccos(-0,5)+2Пn;x=+-2П/3+2Пn.n принадлежит Z.



2cos^2x-2sin2x+1=0
2cos^2x-2sin2x+cos^2x+sin^2x=0
3cos^2x-4sinxcosx+sin^2x=0
делим обе части уравнения на cos^2x, при условии что cosx не равен нулю
3-4tgx+tg^2x=0 заменяем tgx на у
у^2-4у+3=0
решаем квадратное уравнение, находим у, находим tgx, находим х, сравниваем х с условием cosx не равен нулю, пишем ответ

Answer 2

To solve the equation cos(2x) - cos(x) = 0, we can use trigonometric identities.

Step 1: Apply the double-angle formula for cosine: cos(2x) = 2cos^2(x) - 1.

Substituting this into the equation, we get: 2cos^2(x) - 1 - cos(x) = 0.

Step 2: Rearrange the equation: 2cos^2(x) - cos(x) - 1 = 0.

Step 3: Let's make a substitution to simplify the equation. Set y = cos(x), so the equation becomes: 2y^2 - y - 1 = 0.

Step 4: Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we'll use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a).

For the equation 2y^2 - y - 1 = 0, the coefficients are a = 2, b = -1, and c = -1. Plugging these values into the quadratic formula, we have: y = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2)) = (1 ± √(1 + 8)) / 4 = (1 ± √9) / 4 = (1 ± 3) / 4.

Step 5: Solve for y by considering both cases: Case 1: y = (1 + 3) / 4 = 4 / 4 = 1. Case 2: y = (1 - 3) / 4 = -2 / 4 = -1/2.

Step 6: Recall that y = cos(x). Solve for x in each case: Case 1: cos(x) = 1. This occurs when x = 2πn, where n is an integer. Case 2: cos(x) = -1/2. This occurs when x = (2πn ± 2π/3), where n is an integer.

Therefore, the solutions to the equation cos(2x) - cos(x) = 0 are: x = 2πn, where n is an integer. x = (2πn ± 2π/3), where n is an integer.