Question

Помогите решить 2cos²x - 5cos(-3π/2 - x) +1 =0. И объясните пожалуйста

Answer 1

 2cos²x - 5cos(-3π/2 - x) +1 =0
5cos(-3π/2 - x)=5сos(3π/2+x)  в силу четности ф-ии сosx.

по формулам приведения 5сos(3π/2+x) = -5sinx

2cos²x - 5cos(-3π/2 - x) +1 =2cos²x + 5sinx +1 
2cos²x + 5sinx +1=0  2(1-sin²x)+5sinx+1=0
2-2sin²x+5sinx+1=0
-2sin²x+5sinx+3=0
2sin²x-5sinx-3=0  D=25+24=49  √D=7
sinx=1/4[5+7]=3>1
sinx=1/4[5-7]=-1/2

x=(-1)ⁿ(-π/6)+πn   n∈Z

Answer 2

To solve the equation 2cos²x - 5cos(-3π/2 + x) + 1 = 0, we can follow these steps:

Step 1: Simplify the equation Step 2: Substitute cos(-3π/2 + x) with sin(x) Step 3: Solve the resulting quadratic equation

Let's go through each step in detail:

Step 1: Simplify the equation

The equation we have is 2cos²x - 5cos(-3π/2 + x) + 1 = 0. To simplify it, we can use the identity cos(-θ) = cos(θ). So, we have:

2cos²x - 5cos(3π/2 - x) + 1 = 0

Step 2: Substitute cos(-3π/2 + x) with sin(x)

Using the identity cos(θ) = sin(π/2 - θ), we can rewrite the equation as:

2cos²x - 5sin(x) + 1 = 0

Step 3: Solve the resulting quadratic equation

Now, we have a quadratic equation in terms of cos(x). Let's solve it by factoring or using the quadratic formula.

2cos²x - 5sin(x) + 1 = 0

To make it easier, let's substitute cos(x) with y:

2y² - 5sin(x) + 1 = 0

Now, we have a quadratic equation in terms of y. Let's solve it:

2y² - 5sin(x) + 1 = 0

The solutions for y can be found using the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In our case, a = 2, b = -5sin(x), and c = 1. Substituting these values into the quadratic formula, we get:

y = (-(-5sin(x)) ± √((-5sin(x))² - 4(2)(1))) / (2(2))

Simplifying further:

y = (5sin(x) ± √(25sin²(x) - 8)) / 4

Now, we have the solutions for y. To find the solutions for cos(x), we substitute y back into cos(x):

cos(x) = y

cos(x) = (5sin(x) ± √(25sin²(x) - 8)) / 4

Therefore, the solutions for the equation 2cos²x - 5cos(-3π/2 + x) + 1 = 0 are given by the values of x that satisfy the equation:

cos(x) = (5sin(x) ± √(25sin²(x) - 8)) / 4

Please note that this is a general solution, and the specific values of x that satisfy the equation will depend on the range of x and the values of sin(x) that make the equation true.