3cos(2x+pi/7)=√3
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Ответил portechkov
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To solve the equation \(3\cos(2x+\frac{\pi}{7})=\sqrt{3}\), you can start by isolating the cosine term. Divide both sides of the equation by 3:
\[ \cos(2x+\frac{\pi}{7}) = \frac{\sqrt{3}}{3} \]
Now, to find \(x\), take the inverse cosine (arccos) of both sides:
\[ 2x+\frac{\pi}{7} = \arccos\left(\frac{\sqrt{3}}{3}\right) \]
Finally, solve for \(x\). Remember to consider the periodicity of the cosine function, which means there are multiple solutions:
\[ x = \frac{\arccos\left(\frac{\sqrt{3}}{3}\right) - \frac{\pi}{7} + 2n\pi}{2} \]
where \(n\) is an integer.
\[ \cos(2x+\frac{\pi}{7}) = \frac{\sqrt{3}}{3} \]
Now, to find \(x\), take the inverse cosine (arccos) of both sides:
\[ 2x+\frac{\pi}{7} = \arccos\left(\frac{\sqrt{3}}{3}\right) \]
Finally, solve for \(x\). Remember to consider the periodicity of the cosine function, which means there are multiple solutions:
\[ x = \frac{\arccos\left(\frac{\sqrt{3}}{3}\right) - \frac{\pi}{7} + 2n\pi}{2} \]
where \(n\) is an integer.
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