Answer
The solutions to the equation \( \frac{\sin x+\sin 2x}{1+\cos x+\cos 2x}=\tan x \) are all real numbers except \( x = \frac{\pi}{2} + k\pi \), \( x = \frac{2\pi}{3} + 2k\pi \), and \( x = \frac{4\pi}{3} + 2k\pi \) for any integer \( k \).
Solution
Solve the equation \( \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\frac{\sin\left(x\right)+\sin\left(2x\right)}{1+\cos\left(x\right)+\cos\left(2x\right)}=\tan\left(x\right)\)
- step1: Find the domain:
\(\frac{\sin\left(x\right)+\sin\left(2x\right)}{1+\cos\left(x\right)+\cos\left(2x\right)}=\tan\left(x\right),x\neq \left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{2\pi }{3}+2k\pi \\\frac{4\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\)
- step2: Rewrite the expression:
\(\frac{\sin\left(x\right)+2\sin\left(x\right)\cos\left(x\right)}{1+\cos\left(x\right)+2\cos^{2}\left(x\right)-1}=\tan\left(x\right)\)
- step3: Reduce the fraction:
\(\frac{\sin\left(x\right)}{\cos\left(x\right)}=\tan\left(x\right)\)
- step4: Rewrite the expression:
\(\frac{\sin\left(x\right)}{\cos\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}\)
- step5: The statement is true:
\(x \in \mathbb{R}\)
- step6: Check if the solution is in the defined range:
\(x \in \mathbb{R},x\neq \left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{2\pi }{3}+2k\pi \\\frac{4\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\)
- step7: Find the intersection:
\(x\neq \left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{2\pi }{3}+2k\pi \\\frac{4\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\)
The solution to the equation \( \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x \) is \( x \neq \left\{ \begin{array}{l} \frac{\pi }{2}+k\pi \\ \frac{2\pi }{3}+2k\pi \\ \frac{4\pi }{3}+2k\pi \\ \end{array} \right., k \in \mathbb{Z} \).
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