asked by guest
You asked:
Solve the equation
cos ( 2 x ) = 3 2 \cos\left( 2 x \right) = \frac{3}{2} cos ( 2 x ) = 2 3 for the variable
x x x .
MathBot Answer:
There are no real solutions.
The complex solutions are: x = { π n + i log ( 2 ) 2 − i log ( 3 + 5 ) 2 | n ∈ Z } ≈ { 3.1415927 n − 0.48121183 i | n ∈ Z } x = { π n + i log ( 2 ) 2 − i log ( 3 − 5 ) 2 | n ∈ Z } ≈ { 3.1415927 n + 0.48121183 i | n ∈ Z } \begin{aligned}x &= \left\{\pi n + \frac{i \log{\left(2 \right)}}{2} - \frac{i \log{\left(3 + \sqrt{5} \right)}}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n - 0.48121183 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi n + \frac{i \log{\left(2 \right)}}{2} - \frac{i \log{\left(3 - \sqrt{5} \right)}}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 0.48121183 i\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} x x = { πn + 2 i log ( 2 ) − 2 i log ( 3 + 5 ) n ∈ Z } ≈ { 3.1415927 n − 0.48121183 i ∣ n ∈ Z } = { πn + 2 i log ( 2 ) − 2 i log ( 3 − 5 ) n ∈ Z } ≈ { 3.1415927 n + 0.48121183 i ∣ n ∈ Z }
i i i imaginary unit , defined as i 2 = − 1 i^2 = -1 i 2 = − 1
