\text{Solve for }x of 5\cos(x)=5-5\cos(x)

x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}

Solve 5cos(x)=5-5cos(x)

Evaluate

5 cos (x) = 5 - 5 cos (x)

Move the expression to the left side

5 cos (x) - (5 - 5 cos (x)) = 0

Calculate

More Steps

10 cos (x) - 5 = 0

Move the constant to the right-hand side and change its sign

10 cos (x) = 0 + 5

Removing 0 doesn't change the value,so remove it from the expression

10 cos (x) = 5

Divide both sides

\frac{10 cos ( x )}{10} = \frac{5}{10}

Divide the numbers

cos (x) = \frac{5}{10}

Cancel out the common factor 5

cos (x) = \frac{1}{2}

Use the inverse trigonometric function

x = arccos (\frac{1}{2})

Calculate

x = \frac{π}{3} x = \frac{5 π}{3}

Add the period of 2 kπ, k \in Z to find all solutions

x = \frac{π}{3} + 2 kπ, k \in Z x = \frac{5 π}{3} + 2 kπ, k \in Z

Solution

x = {\frac{π}{3} + 2 kπ \frac{5 π}{3} + 2 kπ, k \in Z

Alternative Form

x = {6 0^{\circ} + 36 0^{\circ} k 30 0^{\circ} + 36 0^{\circ} k, k \in Z

Alternative Form

x \approx {1.047198 + 2 kπ 5.235988 + 2 kπ, k \in Z