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Soluțiile ecuațiilor trigonometrice sunt:
6. a) \( x = k\pi \) sau \( x = \frac{\pi}{8} + \frac{k\pi}{4} \), unde \( k \) este un întreg.
b) \( x = \frac{2k\pi}{5} \) sau \( x = \frac{2k\pi}{3} \), unde \( k \) este un întreg.
c) \( x = \frac{3\pi}{16} + \frac{k\pi}{4} \) sau \( x = \frac{\pi}{4} + k\pi \), unde \( k \) este un întreg.
d) \( x = \frac{\pi}{4} \).
e) \( x = \frac{3\pi}{2} + x \) (aici a apărut o eroare).
f) \( x = \frac{3x}{2} + \frac{\pi}{4} \) (aici a apărut o eroare).
7. a) \( x = \frac{k\pi}{3} \) sau \( x = \frac{\pi}{8} + \frac{k\pi}{4} \), unde \( k \) este un întreg.
b) \( x = \frac{2k\pi}{3} \) sau \( x = \pi + 2k\pi \), unde \( k \) este un întreg.
c) \( x = 0 \).
d) \( x = \frac{2k\pi}{9} \), unde \( k \) este un întreg.
e) \( x = 0 \).
f) \( x = \frac{\pi}{8} + \frac{k\pi}{4} \) sau \( x = \frac{\pi}{4} + \frac{k\pi}{2} \), unde \( k \) este un întreg.
Dacă ai nevoie de mai multe detalii sau de ajutor cu anumite ecuații, te rog să mă întrebi!
Solución
Solve the equation by following steps:
- step0: Solve for \(x\):
\(2x-x=0\)
- step1: Subtract the terms:
\(x=0\)
Solve the equation \( \text{tg}(4x) - \text{tg}(2x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(4x-2x=0\)
- step1: Subtract the terms:
\(2x=0\)
- step2: Rewrite the expression:
\(x=0\)
Solve the equation \( \cos(4x) - \cos(5x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\cos\left(4x\right)-\cos\left(5x\right)=0\)
- step1: Transform the expression:
\(-2\sin\left(\frac{9x}{2}\right)\sin\left(-\frac{x}{2}\right)=0\)
- step2: Elimination the left coefficient:
\(\sin\left(\frac{9x}{2}\right)\sin\left(-\frac{x}{2}\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\sin\left(\frac{9x}{2}\right)=0\\&\sin\left(-\frac{x}{2}\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{2k\pi }{9},k \in \mathbb{Z}\\&x=2k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\frac{2k\pi }{9},k \in \mathbb{Z}\)
Solve the equation \( \sin(5x) - \sin(3x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\sin\left(5x\right)-\sin\left(3x\right)=0\)
- step1: Transform the expression:
\(2\cos\left(4x\right)\sin\left(x\right)=0\)
- step2: Elimination the left coefficient:
\(\cos\left(4x\right)\sin\left(x\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\cos\left(4x\right)=0\\&\sin\left(x\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{\pi }{8}+\frac{k\pi }{4},k \in \mathbb{Z}\\&x=k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\left\{ \begin{array}{l}k\pi \\\frac{\pi }{8}+\frac{k\pi }{4}\end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( \cos(5x) + \sin(3x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\cos\left(5x\right)+\sin\left(3x\right)=0\)
- step1: Transform the expression:
\(2\sin\left(\frac{16x+\pi }{4}\right)\cos\left(\frac{4x+\pi }{4}\right)=0\)
- step2: Elimination the left coefficient:
\(\sin\left(\frac{16x+\pi }{4}\right)\cos\left(\frac{4x+\pi }{4}\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\sin\left(\frac{16x+\pi }{4}\right)=0\\&\cos\left(\frac{4x+\pi }{4}\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{3\pi }{16}+\frac{k\pi }{4},k \in \mathbb{Z}\\&x=\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\left\{ \begin{array}{l}\frac{3\pi }{16}+\frac{k\pi }{4}\\\frac{\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( \text{tg}(4x) - \text{ctg}(3x + \frac{\pi}{4}) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(4x-\left(3x+\frac{\pi }{4}\right)=0\)
- step1: Subtract the terms:
\(x-\frac{\pi }{4}=0\)
- step2: Move the constant to the right side:
\(x=0+\frac{\pi }{4}\)
- step3: Add the terms:
\(x=\frac{\pi }{4}\)
Solve the equation \( \cos(x) - \cos(4x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\cos\left(x\right)-\cos\left(4x\right)=0\)
- step1: Transform the expression:
\(-2\sin\left(\frac{5x}{2}\right)\sin\left(-\frac{3x}{2}\right)=0\)
- step2: Elimination the left coefficient:
\(\sin\left(\frac{5x}{2}\right)\sin\left(-\frac{3x}{2}\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\sin\left(\frac{5x}{2}\right)=0\\&\sin\left(-\frac{3x}{2}\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{2k\pi }{5},k \in \mathbb{Z}\\&x=\frac{2k\pi }{3},k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\left\{ \begin{array}{l}\frac{2k\pi }{5}\\\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( \sin(x) - \sin(7x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\sin\left(x\right)-\sin\left(7x\right)=0\)
- step1: Transform the expression:
\(2\cos\left(4x\right)\sin\left(-3x\right)=0\)
- step2: Elimination the left coefficient:
\(\cos\left(4x\right)\sin\left(-3x\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\cos\left(4x\right)=0\\&\sin\left(-3x\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{\pi }{8}+\frac{k\pi }{4},k \in \mathbb{Z}\\&x=\frac{k\pi }{3},k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\left\{ \begin{array}{l}\frac{k\pi }{3}\\\frac{\pi }{8}+\frac{k\pi }{4}\end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( -\cos(2x) - \cos(6x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(-\cos\left(2x\right)-\cos\left(6x\right)=0\)
- step1: Transform the expression:
\(-2\cos\left(x+3x\right)\cos\left(-2x\right)=0\)
- step2: Elimination the left coefficient:
\(\cos\left(x+3x\right)\cos\left(-2x\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\cos\left(x+3x\right)=0\\&\cos\left(-2x\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{\pi }{8}+\frac{k\pi }{4},k \in \mathbb{Z}\\&x=\frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\left\{ \begin{array}{l}\frac{\pi }{8}+\frac{k\pi }{4}\\\frac{\pi }{4}+\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( -\sin(2x) - \sin(x) = 0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(-\sin\left(2x\right)-\sin\left(x\right)=0\)
- step1: Transform the expression:
\(-2\sin\left(\frac{2x+x}{2}\right)\cos\left(\frac{x}{2}\right)=0\)
- step2: Elimination the left coefficient:
\(\sin\left(\frac{2x+x}{2}\right)\cos\left(\frac{x}{2}\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&\sin\left(\frac{2x+x}{2}\right)=0\\&\cos\left(\frac{x}{2}\right)=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=\frac{2k\pi }{3},k \in \mathbb{Z}\\&x=\pi +2k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Find the union:
\(x=\left\{ \begin{array}{l}\frac{2k\pi }{3}\\\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\)
Iată soluțiile pentru ecuațiile trigonometrice date:
### 6. Rezolvarea ecuațiilor trigonometrice:
a) \( \sin 5x = \sin 3x \)
Soluția este:
\[
x = \left\{
\begin{array}{l}
k\pi \\
\frac{\pi}{8} + \frac{k\pi}{4}
\end{array}
\right., k \in \mathbb{Z}
\]
b) \( \cos x = \cos 4x \)
Soluția este:
\[
x = \left\{
\begin{array}{l}
\frac{2k\pi}{5} \\
\frac{2k\pi}{3}
\end{array}
\right., k \in \mathbb{Z}
\]
c) \( \cos 5x = -\sin 3x \)
Soluția este:
\[
x = \left\{
\begin{array}{l}
\frac{3\pi}{16} + \frac{k\pi}{4} \\
\frac{\pi}{4} + k\pi
\end{array}
\right., k \in \mathbb{Z}
\]
d) \( \tan 4x = \cot\left(3x + \frac{\pi}{4}\right) \)
Soluția este:
\[
x = \frac{\pi}{4}
\]
e) \( \sin\left(\frac{3\pi}{2} + x\right) + \cos 3x = 0 \)
Aici a apărut o eroare în procesarea ecuației. Voi încerca să o rezolv din nou.
f) \( \cos\left(\frac{3x}{2} + \frac{\pi}{4}\right) + \sin\left(\frac{2x}{3} - \frac{\pi}{4}\right) = 0 \)
Aici a apărut o eroare în procesarea ecuației. Voi încerca să o rezolv din nou.
### 7. Rezolvarea ecuațiilor trigonometrice:
a) \( \sin x = \sin 7x \)
Soluția este:
\[
x = \left\{
\begin{array}{l}
\frac{k\pi}{3} \\
\frac{\pi}{8} + \frac{k\pi}{4}
\end{array}
\right., k \in \mathbb{Z}
\]
b) \( -\sin 2x = \sin x \)
Soluția este:
\[
x = \left\{
\begin{array}{l}
\frac{2k\pi}{3} \\
\pi + 2k\pi
\end{array}
\right., k \in \mathbb{Z}
\]
c) \( \tan 4x = \tan 2x \)
Soluția este:
\[
x = 0
\]
d) \( \cos 4x = \cos 5x \)
Soluția este:
\[
x = \frac{2k\pi}{9}, k \in \mathbb{Z}
\]
e) \( -\cos 2x = -\cos 6x \)
Soluția este:
\[
x = 0
\]
f) \( \cot 2x = \cot x \)
Soluția este:
\[
x = \left\{
\begin{array}{l}
\frac{\pi}{8} + \frac{k\pi}{4} \\
\frac{\pi}{4} + \frac{k\pi}{2}
\end{array}
\right., k \in \mathbb{Z}
\]
Dacă ai nevoie de detalii suplimentare sau de clarificări pentru anumite ecuații, te rog să îmi spui!
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