Question

2cos2x=7cosx

Answer 1

2cos2x=7cosx 2cos2x-7cosx=0 2(cosx)^2-2(sinx)^2-7cosx=0 2(cosx)^2-2(1-(cosx)^2)-7cosx=0 2(cosx)^2-2+2(cosx)^2-7cosx=0 4(cosx)^2-7cosx-2=0 t=cosx 4t^2-7t-2=0 D=(-7)^2-4*4*(-2)=49+32=81 t1=(7+9):8=2         t2=(7-9):8=-1/4 cosx=2                                                         cosx=-1/4  корней нет, т.к. |cosx|<1                           x=пи-arccos(1/4)+2пиэн, эн принадлежит Z

Answer 2

To solve the equation 2cos(2x) = 7cos(x), you can use the double-angle formula for cosine, which states:

cos(2x) = 2cos^2(x) - 1

So, you can substitute this expression into your equation:

2(2cos^2(x) - 1) = 7cos(x)

Now, simplify and rearrange the equation:

4cos^2(x) - 2 = 7cos(x)

Now, let's make a substitution to simplify further. Let y = cos(x), so the equation becomes:

4y^2 - 2 = 7y

Rearrange the terms:

4y^2 - 7y - 2 = 0

Now, you have a quadratic equation in terms of y. You can solve this quadratic equation using the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 4, b = -7, and c = -2. Plug these values into the formula:

y = (7 ± √((-7)² - 4(4)(-2))) / (2(4))

Now, calculate the discriminant (the value inside the square root):

Discriminant = (-7)² - 4(4)(-2) = 49 + 32 = 81

Now, substitute the discriminant back into the formula:

y = (7 ± √81) / (2(4))

y = (7 ± 9) / 8

Now, calculate both possible values for y:

1. y = (7 + 9) / 8 = 16 / 8 = 2
2. y = (7 - 9) / 8 = -2 / 8 = -1/4

Now that you have the possible values for y, which represents cos(x), you can find the corresponding values for x. Remember that the range of the cosine function is between -1 and 1, so the solution for y = -1/4 is extraneous, and you can discard it.

So, the valid solution for cos(x) = 2 is:

cos(x) = 2

However, there is no real number x for which cos(x) equals 2, as the range of the cosine function is [-1, 1]. Therefore, there are no real solutions to the original equation 2cos(2x) = 7cos(x).