1.3 $$ \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}= an x $$

Question

1.3 $$ \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=	an x $$

UpStudy AI · Accepted Answer

Solve the equation $$ \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=	an x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{\sin\left(xight)+\sin\left(2xight)}{1+\cos\left(xight)+\cos\left(2xight)}=	an\left(xight)$$
- step1: Find the domain:
  $$\frac{\sin\left(xight)+\sin\left(2xight)}{1+\cos\left(xight)+\cos\left(2xight)}=	an\left(xight),x
eq \left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\frac{2\pi }{3}+2k\pi \\frac{4\pi }{3}+2k\pi \end{array}ight.,k \in \mathbb{Z}$$
- step2: Rewrite the expression:
  $$\frac{\sin\left(xight)+2\sin\left(xight)\cos\left(xight)}{1+\cos\left(xight)+2\cos^{2}\left(xight)-1}=	an\left(xight)$$
- step3: Reduce the fraction:
  $$\frac{\sin\left(xight)}{\cos\left(xight)}=	an\left(xight)$$
- step4: Rewrite the expression:
  $$\frac{\sin\left(xight)}{\cos\left(xight)}=\frac{\sin\left(xight)}{\cos\left(xight)}$$
- step5: The statement is true:
  $$x \in \mathbb{R}$$
- step6: Check if the solution is in the defined range:
  $$x \in \mathbb{R},x
eq \left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\frac{2\pi }{3}+2k\pi \\frac{4\pi }{3}+2k\pi \end{array}ight.,k \in \mathbb{Z}$$
- step7: Find the intersection:
  $$x
eq \left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\frac{2\pi }{3}+2k\pi \\frac{4\pi }{3}+2k\pi \end{array}ight.,k \in \mathbb{Z}$$
The solution to the equation $$ \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=	an x $$ is $$ x 
eq \left\{ \begin{array}{l} \frac{\pi }{2}+k\pi \ \frac{2\pi }{3}+2k\pi \ \frac{4\pi }{3}+2k\pi \ \end{array} ight., k \in \mathbb{Z} $$.
The solutions to the equation $$ \frac{\sin x+\sin 2x}{1+\cos x+\cos 2x}=	an x $$ 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 $$ k $$.

1.3 \( \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x \)

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