Question

0,5cos(2x)-cos(x)+2cos^3(x)=0

Answer 1

cos2x=2cos²x-1

0,5·(2cos²x-1)-cosx+2cos³x=0

cos²x-0,5-cosx+2cos³x=0

Раскладываем левую часть на множители способом группировки:

(2cos³x-cosx)+(cos²x-0,5)=0

2cosx·(cos²x-0,5)+(cos²x-0,5)=0

(cos²x-0,5)·(2cosx+1)=0

cos²x-0,5=0   или  2cosx+1=0

cos²x-0,5=0  

cosx=±√2/2

cosx=√2/2  или  cosx=-√2/2

x=(±π/4)+2πn  или x=(±3π/4)++2πm, n,m∈Z

2cosx+1=0

cosx=-1/2

x=(±2π/3)+2πk, k∈Z

О т в е т. (±π/4)+2πn ; (±3π/4)++2πm ; (±2π/3)+2πk, n,m, k∈Z

Answer 2

To solve the trigonometric equation 0.5cos(2x) - cos(x) + 2cos^3(x) = 0, we'll first use some trigonometric identities to simplify it, and then find the solutions. Here's the step-by-step process:

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

Step 2: Substitute the double-angle identity into the equation. 0.5(2cos^2(x) - 1) - cos(x) + 2cos^3(x) = 0

Step 3: Simplify the equation. cos^2(x) - cos(x) + 2cos^3(x) - 0.5 = 0

Step 4: Factor out common terms. cos^2(x) + 2cos^3(x) - cos(x) - 0.5 = 0

Step 5: Group the terms and factor by cos(x). cos(x)(cos^2(x) + 2cos^2(x) - 1) - 0.5 = 0

Step 6: Combine like terms. cos(x)(3cos^2(x) - 1) - 0.5 = 0

Now, we have a quadratic equation in terms of cos(x): 3cos^3(x) - cos(x) - 0.5 = 0

Let's solve this quadratic equation for cos(x). To do this, we can use the quadratic formula: For any quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by: x = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 3, b = -1, and c = -0.5.

Step 7: Substitute the values into the quadratic formula. cos(x) = [1 ± √((-1)^2 - 4 * 3 * (-0.5))] / (2 * 3)

Step 8: Simplify under the square root. cos(x) = [1 ± √(1 + 6)] / 6 cos(x) = [1 ± √7] / 6

So the two possible solutions for cos(x) are:

1. cos(x) = (1 + √7) / 6
2. cos(x) = (1 - √7) / 6

Step 9: Find the values of x. To find the values of x, we need to use the inverse cosine function (arccosine). In general, for any real number "a," the solutions for x in cos(x) = a are given by: x = arccos(a) + 2πn or x = -arccos(a) + 2πn, where "n" is an integer.

Using this formula, we can find the values of x corresponding to the two solutions for cos(x) obtained above.

For cos(x) = (1 + √7) / 6: x = arccos((1 + √7) / 6) + 2πn or x = -arccos((1 + √7) / 6) + 2πn

And for cos(x) = (1 - √7) / 6: x = arccos((1 - √7) / 6) + 2πn or x = -arccos((1 - √7) / 6) + 2πn

Keep in mind that "n" can be any integer, as it represents the periodic nature of the trigonometric functions.

These are the solutions for the given trigonometric equation.