(7) $$ an \alpha=\cos \alpha $$

Question

(7) $$ 	an \alpha=\cos \alpha $$

UpStudy AI · Accepted Answer

Solve the equation $$ \tan(\alpha)=\cos(\alpha) $$.
Solve the equation by following steps:
- step0: Solve for $$\alpha$$:
  $$\tan\left(\alpha \right)=\cos\left(\alpha \right)$$
- step1: Find the domain:
  $$\tan\left(\alpha \right)=\cos\left(\alpha \right),\alpha \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step2: Rewrite the expression:
  $$\frac{\sin\left(\alpha \right)}{\cos\left(\alpha \right)}=\cos\left(\alpha \right)$$
- step3: Cross multiply:
  $$\sin\left(\alpha \right)=\cos\left(\alpha \right)\cos\left(\alpha \right)$$
- step4: Simplify the equation:
  $$\sin\left(\alpha \right)=\cos^{2}\left(\alpha \right)$$
- step5: Rewrite the expression:
  $$\sin\left(\alpha \right)=1-\sin^{2}\left(\alpha \right)$$
- step6: Move the expression to the left side:
  $$\sin\left(\alpha \right)-\left(1-\sin^{2}\left(\alpha \right)\right)=0$$
- step7: Calculate:
  $$\sin\left(\alpha \right)-1+\sin^{2}\left(\alpha \right)=0$$
- step8: Rewrite in standard form:
  $$\sin^{2}\left(\alpha \right)+\sin\left(\alpha \right)-1=0$$
- step9: Solve using the quadratic formula:
  $$\sin\left(\alpha \right)=\frac{-1\pm \sqrt{1^{2}-4\left(-1\right)}}{2}$$
- step10: Simplify the expression:
  $$\sin\left(\alpha \right)=\frac{-1\pm \sqrt{5}}{2}$$
- step11: Separate into possible cases:
  $$\begin{align}&\sin\left(\alpha \right)=\frac{-1+\sqrt{5}}{2}\&\sin\left(\alpha \right)=\frac{-1-\sqrt{5}}{2}\end{align}$$
- step12: Rewrite the fraction:
  $$\begin{align}&\sin\left(\alpha \right)=\frac{-1+\sqrt{5}}{2}\&\sin\left(\alpha \right)=-\frac{1+\sqrt{5}}{2}\end{align}$$
- step13: Rearrange the terms:
  $$\begin{align}&\sin\left(\alpha \right)=\frac{-1+\sqrt{5}}{2}\&\alpha  \notin \mathbb{R}\end{align}$$
- step14: Calculate:
  $$\begin{align}&\alpha =\left\{ \begin{array}{l}\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+2k\pi \-\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\&\alpha  \notin \mathbb{R}\end{align}$$
- step15: Find the union:
  $$\alpha =\left\{ \begin{array}{l}\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+2k\pi \-\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}$$
- step16: Check if the solution is in the defined range:
  $$\alpha =\left\{ \begin{array}{l}\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+2k\pi \-\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z},\alpha \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step17: Find the intersection:
  $$\alpha =\left\{ \begin{array}{l}\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+2k\pi \-\arcsin\left(\frac{-1+\sqrt{5}}{2}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}$$
The solution to the equation $$ \tan \alpha = \cos \alpha $$ is given by:

$$ \alpha = \left\{ \begin{array}{l} \arcsin\left(\frac{-1+\sqrt{5}}{2}\right) + 2k\pi \ -\arcsin\left(\frac{-1+\sqrt{5}}{2}\right) + \pi + 2k\pi \end{array} \right., k \in \mathbb{Z} $$

where $$ k $$ is an integer.
The solutions to the equation $$ \tan \alpha = \cos \alpha $$ are: $$ \alpha = \arcsin\left(\frac{-1+\sqrt{5}}{2}\right) + 2k\pi $$ $$ \alpha = -\arcsin\left(\frac{-1+\sqrt{5}}{2}\right) + \pi + 2k\pi $$ where $$ k $$ is any integer.

(7) \( \tan \alpha=\cos \alpha \)

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