Question
\( \tan 10 x=\cot 5 x \) then \( x= \) ?
Ask by Bowman Mcguire. in Bangladesh
Dec 03,2024
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( x=\frac{\pi}{30}+\frac{k\pi}{15} \)
Solution
Solve the equation \( \tan(10x)=\cot(5x) \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\tan\left(10x\right)=\cot\left(5x\right)\)
- step1: Find the domain:
\(\tan\left(10x\right)=\cot\left(5x\right),x\neq \left\{ \begin{array}{l}\frac{k\pi }{5}\\\frac{\pi }{20}+\frac{k\pi }{10}\end{array}\right.,k \in \mathbb{Z}\)
- step2: Rewrite the expression:
\(\tan\left(10x\right)=\frac{\cos\left(5x\right)}{\sin\left(5x\right)}\)
- step3: Rewrite the expression:
\(\frac{2\tan\left(5x\right)}{1-\tan^{2}\left(5x\right)}=\frac{\cos\left(5x\right)}{\sin\left(5x\right)}\)
- step4: Rewrite the expression:
\(\frac{2\times \frac{\sin\left(5x\right)}{\cos\left(5x\right)}}{1-\left(\frac{\sin\left(5x\right)}{\cos\left(5x\right)}\right)^{2}}=\frac{\cos\left(5x\right)}{\sin\left(5x\right)}\)
- step5: Expand the expression:
\(\frac{\frac{2\sin\left(5x\right)}{\cos\left(5x\right)}}{1-\frac{\sin^{2}\left(5x\right)}{\cos^{2}\left(5x\right)}}=\frac{\cos\left(5x\right)}{\sin\left(5x\right)}\)
- step6: Calculate:
\(\frac{2\sin\left(5x\right)\cos\left(5x\right)}{\cos^{2}\left(5x\right)-\sin^{2}\left(5x\right)}=\frac{\cos\left(5x\right)}{\sin\left(5x\right)}\)
- step7: Cross multiply:
\(2\sin\left(5x\right)\cos\left(5x\right)\sin\left(5x\right)=\left(\cos^{2}\left(5x\right)-\sin^{2}\left(5x\right)\right)\cos\left(5x\right)\)
- step8: Simplify the equation:
\(2\sin^{2}\left(5x\right)\cos\left(5x\right)=\left(\cos^{2}\left(5x\right)-\sin^{2}\left(5x\right)\right)\cos\left(5x\right)\)
- step9: Rewrite the expression:
\(2\cos\left(5x\right)-2\cos^{3}\left(5x\right)=2\cos^{3}\left(5x\right)-\cos\left(5x\right)\)
- step10: Move the expression to the left side:
\(2\cos\left(5x\right)-2\cos^{3}\left(5x\right)-\left(2\cos^{3}\left(5x\right)-\cos\left(5x\right)\right)=0\)
- step11: Calculate:
\(3\cos\left(5x\right)-4\cos^{3}\left(5x\right)=0\)
- step12: Factor the expression:
\(\cos\left(5x\right)\left(3-4\cos^{2}\left(5x\right)\right)=0\)
- step13: Separate into possible cases:
\(\begin{align}&\cos\left(5x\right)=0\\&3-4\cos^{2}\left(5x\right)=0\end{align}\)
- step14: Solve the equation:
\(\begin{align}&x=\frac{\pi }{10}+\frac{k\pi }{5},k \in \mathbb{Z}\\&x=\left\{ \begin{array}{l}\frac{\pi }{30}+\frac{k\pi }{5}\\\frac{\pi }{6}+\frac{k\pi }{5}\end{array}\right.,k \in \mathbb{Z}\end{align}\)
- step15: Find the union:
\(x=\frac{\pi }{30}+\frac{k\pi }{15},k \in \mathbb{Z}\)
- step16: Check if the solution is in the defined range:
\(x=\frac{\pi }{30}+\frac{k\pi }{15},k \in \mathbb{Z},x\neq \left\{ \begin{array}{l}\frac{k\pi }{5}\\\frac{\pi }{20}+\frac{k\pi }{10}\end{array}\right.,k \in \mathbb{Z}\)
- step17: Find the intersection:
\(x=\frac{\pi }{30}+\frac{k\pi }{15},k \in \mathbb{Z}\)
The solution to the equation \( \tan(10x)=\cot(5x) \) is \( x=\frac{\pi}{30}+\frac{k\pi}{15} \), where \( k \) is an integer.
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Simplify this solution Bonus Knowledge
To solve the equation \( \tan (10x) = \cot (5x) \), we can first use the identity \( \cot (5x) = \frac{1}{\tan (5x)} \). Thus, we rewrite the equation as: \[ \tan (10x) = \frac{1}{\tan (5x)} \] This implies: \[ \tan (10x) \tan (5x) = 1 \] Using the identity \( \tan (A) \tan (B) = 1 \) means that \( A + B = \frac{\pi}{2} + n\pi \) for any integer \( n \). Here, we set \( A = 10x \) and \( B = 5x \): \[ 10x + 5x = \frac{\pi}{2} + n\pi \] Simplifying this gives: \[ 15x = \frac{\pi}{2} + n\pi \] Dividing through by 15: \[ x = \frac{\pi}{30} + \frac{n\pi}{15} \] Thus, the general solution for \( x \) is: \[ x = \frac{\pi}{30} + \frac{n\pi}{15} \] where \( n \) is any integer.
