cos6x + cos2x = 0

asked by guest
on Dec 28, 2024 at 9:16 am



You asked:

Solve the equation \(\cos\left( 6 x \right) + \cos\left( 2 x \right) = 0\) for the variable \(x\).

MathBot Answer:

The real solutions are: \[\begin{aligned}x &= \left\{\pi \left(\frac{1}{4} + n\right)\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 0.78539816\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi \left(\frac{1}{8} + n\right)\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 0.39269908\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(3 + 4 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 2.3561945\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(5 + 8 n\right)}{8}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 1.9634954\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(3 + 8 n\right)}{8}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 1.1780972\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(7 + 8 n\right)}{8}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 2.7488936\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

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