Question

Решите: cos^2x=1/2(1-cosx)

Answer 1

$Cos^{2}x=\frac{1-Cosx}{2}\\\\Cos^{2}x-\frac{1-Cosx}{2}=0 \\\\2Cos^{2}x-1+Cosx=0\\\\2Cos^{2}x+Cosx-1=0\\\\Cosx=m,-1 \leq m \leq 1 \\\\2m^{2}+m-1=0\\\\D=1^{2} -4*2*(-1)=1+8=9=3^{2}\\\\m_{1}=\frac{-1+3}{4}=\frac{1}{2} \\\\m_{2} =\frac{-1-3}{4}=-1\\\\Cosx=\frac{1}{2}\\\\x=+-arcCos\frac{1}{2}+2\pi n,nez\\\\x=+-\frac{\pi }{3}+2\pi n,nez\\\\Cosx=-1\\\\x=\pi +2\pi n,nez$

Answer 2

To solve the equation cos^2x = 1/2(1 - cosx), we'll first rewrite it to make it easier to work with. Let's move all terms to one side of the equation:

cos^2x - 1/2(1 - cosx) = 0

Now, let's simplify the equation:

cos^2x - 1/2 + 1/2 * cosx = 0

To proceed, let's make a substitution to simplify further. Let u = cos(x):

u^2 - 1/2 + 1/2 * u = 0

Now, we have a quadratic equation in terms of u. To solve it, we'll set the equation equal to zero and factor it (if possible):

u^2 + 1/2 * u - 1/2 = 0

To factor the equation, we look for two numbers that multiply to -1/2 and add up to 1/2. The numbers are 1 and -1/2:

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

Now, we can solve for u:

u + 1 = 0 or u - 1/2 = 0

If u + 1 = 0, then u = -1 If u - 1/2 = 0, then u = 1/2

Now, we need to convert back from u to x:

1. If u = cos(x) = -1: x = arccos(-1) = π (plus any integer multiple of 2π since cos has a period of 2π)

2. If u = cos(x) = 1/2: x = arccos(1/2) = π/3 (plus any integer multiple of 2π or 4π since cos has a period of 2π)

So, the solutions to the equation cos^2x = 1/2(1 - cosx) are x = π + 2nπ and x = π/3 + 2mπ or 4mπ, where n and m are integers.