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Here are the general solutions for the given equations:
1. \( \sin(x - 16^\circ) = 0.616 \)
\[
x \approx -0.330808 + k\pi \quad \text{or} \quad x \approx 0.330808 + k\pi, \quad k \in \mathbb{Z}
\]
2. \( \cos(2x) = 0.789 \)
\[
x \approx -0.564703 + \frac{2k\pi}{3} \quad \text{or} \quad x \approx 0.564703 + \frac{2k\pi}{3}, \quad k \in \mathbb{Z}
\]
3. \( \cos(3x) = -0.123 \)
\[
x \approx -0.564703 + \frac{2k\pi}{3} \quad \text{or} \quad x \approx 0.564703 + \frac{2k\pi}{3}, \quad k \in \mathbb{Z}
\]
4. \( \tan(x + 56^\circ) = 7.56 \)
\[
x \approx -54.560715 + k\pi, \quad k \in \mathbb{Z}
\]
5. \( \tan\left(\frac{x}{2}\right) = -1.421 \)
\[
x \approx -1.915143 + 2k\pi, \quad k \in \mathbb{Z}
\]
6. \( \sin(2x + 44^\circ) = -0.708 \)
\[
x \approx -22.393331 + k\pi \quad \text{or} \quad x \approx -20.035873 + k\pi, \quad k \in \mathbb{Z}
\]
7. \( 3\cos(x - 15^\circ) + 1 = -0.456 \)
\[
x = -\arccos\left(-\frac{182}{375}\right) + 15 + 2k\pi \quad \text{or} \quad x = \arccos\left(-\frac{182}{375}\right) + 15 + 2k\pi, \quad k \in \mathbb{Z}
\]
8. \( 2\tan(2x - 10^\circ) = 10.67 \)
\[
x = \frac{\arctan\left(\frac{1067}{200}\right) + 10}{2} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}
\]
These solutions provide the general forms for each equation, where \( k \) is any integer.
Solución
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\tan\left(x+56\right)=7.56\)
- step1: Find the domain:
\(\tan\left(x+56\right)=7.56,x\neq \frac{37\pi -112}{2}+k\pi ,k \in \mathbb{Z}\)
- step2: Use the inverse trigonometric function:
\(x+56=\arctan\left(7.56\right)\)
- step3: Add the period:
\(x+56=\arctan\left(7.56\right)+k\pi ,k \in \mathbb{Z}\)
- step4: Solve the equation:
\(x=\arctan\left(7.56\right)-56+k\pi ,k \in \mathbb{Z}\)
- step5: Check if the solution is in the defined range:
\(x=\arctan\left(7.56\right)-56+k\pi ,k \in \mathbb{Z},x\neq \frac{37\pi -112}{2}+k\pi ,k \in \mathbb{Z}\)
- step6: Find the intersection:
\(x=\arctan\left(7.56\right)-56+k\pi ,k \in \mathbb{Z}\)
- step7: Simplify:
\(x\approx -54.560715+k\pi ,k \in \mathbb{Z}\)
Solve the equation \( 2\tan(2x-10^{\text{°}})=10.67 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(2\tan\left(2x-10\right)=10.67\)
- step1: Find the domain:
\(2\tan\left(2x-10\right)=10.67,x\neq \frac{\pi +20}{4}+\frac{k\pi }{2},k \in \mathbb{Z}\)
- step2: Multiply both sides:
\(2\tan\left(2x-10\right)\times \frac{1}{2}=10.67\times \frac{1}{2}\)
- step3: Calculate:
\(\tan\left(2x-10\right)=10.67\times \frac{1}{2}\)
- step4: Calculate:
\(\tan\left(2x-10\right)=\frac{1067}{200}\)
- step5: Use the inverse trigonometric function:
\(2x-10=\arctan\left(\frac{1067}{200}\right)\)
- step6: Add the period:
\(2x-10=\arctan\left(\frac{1067}{200}\right)+k\pi ,k \in \mathbb{Z}\)
- step7: Solve the equation:
\(x=\frac{\arctan\left(\frac{1067}{200}\right)+10}{2}+\frac{k\pi }{2},k \in \mathbb{Z}\)
- step8: Check if the solution is in the defined range:
\(x=\frac{\arctan\left(\frac{1067}{200}\right)+10}{2}+\frac{k\pi }{2},k \in \mathbb{Z},x\neq \frac{\pi +20}{4}+\frac{k\pi }{2},k \in \mathbb{Z}\)
- step9: Find the intersection:
\(x=\frac{\arctan\left(\frac{1067}{200}\right)+10}{2}+\frac{k\pi }{2},k \in \mathbb{Z}\)
Solve the equation \( \cos(2x)=0.789 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\cos\left(2x\right)=0.789\)
- step1: Use the inverse trigonometric function:
\(2x=\arccos\left(0.789\right)\)
- step2: Calculate:
\(\begin{align}&2x=-\arccos\left(0.789\right)\\&2x=\arccos\left(0.789\right)\end{align}\)
- step3: Add the period:
\(\begin{align}&2x=-\arccos\left(0.789\right)+2k\pi ,k \in \mathbb{Z}\\&2x=\arccos\left(0.789\right)+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step4: Calculate:
\(\begin{align}&x=-\frac{\arccos\left(0.789\right)}{2}+k\pi ,k \in \mathbb{Z}\\&2x=\arccos\left(0.789\right)+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Calculate:
\(\begin{align}&x=-\frac{\arccos\left(0.789\right)}{2}+k\pi ,k \in \mathbb{Z}\\&x=\frac{\arccos\left(0.789\right)}{2}+k\pi ,k \in \mathbb{Z}\end{align}\)
- step6: Find the union:
\(x=\left\{ \begin{array}{l}-\frac{\arccos\left(0.789\right)}{2}+k\pi \\\frac{\arccos\left(0.789\right)}{2}+k\pi \end{array}\right.,k \in \mathbb{Z}\)
- step7: Simplify:
\(x\approx \left\{ \begin{array}{l}-0.330808+k\pi \\0.330808+k\pi \end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( \sin(2x+44^{\text{°}})=-0.708 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\sin\left(2x+44\right)=-0.708\)
- step1: Use the inverse trigonometric function:
\(2x+44=\arcsin\left(-0.708\right)\)
- step2: Calculate:
\(\begin{align}&2x+44=-\arcsin\left(0.708\right)\\&2x+44=\arcsin\left(0.708\right)+\pi \end{align}\)
- step3: Add the period:
\(\begin{align}&2x+44=-\arcsin\left(0.708\right)+2k\pi ,k \in \mathbb{Z}\\&2x+44=\arcsin\left(0.708\right)+\pi +2k\pi ,k \in \mathbb{Z}\end{align}\)
- step4: Calculate:
\(\begin{align}&x=-\frac{\arcsin\left(0.708\right)+44}{2}+k\pi ,k \in \mathbb{Z}\\&2x+44=\arcsin\left(0.708\right)+\pi +2k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Calculate:
\(\begin{align}&x=-\frac{\arcsin\left(0.708\right)+44}{2}+k\pi ,k \in \mathbb{Z}\\&x=\frac{\arcsin\left(0.708\right)+\pi -44}{2}+k\pi ,k \in \mathbb{Z}\end{align}\)
- step6: Find the union:
\(x=\left\{ \begin{array}{l}-\frac{\arcsin\left(0.708\right)+44}{2}+k\pi \\\frac{\arcsin\left(0.708\right)+\pi -44}{2}+k\pi \end{array}\right.,k \in \mathbb{Z}\)
- step7: Simplify:
\(x\approx \left\{ \begin{array}{l}-22.393331+k\pi \\-20.035873+k\pi \end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( \tan\frac{x}{2}=-1.421 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\tan\left(\frac{x}{2}\right)=-1.421\)
- step1: Find the domain:
\(\tan\left(\frac{x}{2}\right)=-1.421,x\neq \pi +2k\pi ,k \in \mathbb{Z}\)
- step2: Use the inverse trigonometric function:
\(\frac{x}{2}=\arctan\left(-1.421\right)\)
- step3: Add the period:
\(\frac{x}{2}=\arctan\left(-1.421\right)+k\pi ,k \in \mathbb{Z}\)
- step4: Solve the equation:
\(x=2\arctan\left(-1.421\right)+2k\pi ,k \in \mathbb{Z}\)
- step5: Check if the solution is in the defined range:
\(x=2\arctan\left(-1.421\right)+2k\pi ,k \in \mathbb{Z},x\neq \pi +2k\pi ,k \in \mathbb{Z}\)
- step6: Find the intersection:
\(x=2\arctan\left(-1.421\right)+2k\pi ,k \in \mathbb{Z}\)
- step7: Simplify:
\(x\approx -1.915143+2k\pi ,k \in \mathbb{Z}\)
Solve the equation \( \cos(3x)=-0.123 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\cos\left(3x\right)=-0.123\)
- step1: Use the inverse trigonometric function:
\(3x=\arccos\left(-0.123\right)\)
- step2: Calculate:
\(\begin{align}&3x=-\arccos\left(-0.123\right)\\&3x=\arccos\left(-0.123\right)\end{align}\)
- step3: Add the period:
\(\begin{align}&3x=-\arccos\left(-0.123\right)+2k\pi ,k \in \mathbb{Z}\\&3x=\arccos\left(-0.123\right)+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step4: Calculate:
\(\begin{align}&x=-\frac{\arccos\left(-0.123\right)}{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\\&3x=\arccos\left(-0.123\right)+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step5: Calculate:
\(\begin{align}&x=-\frac{\arccos\left(-0.123\right)}{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\\&x=\frac{\arccos\left(-0.123\right)}{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\end{align}\)
- step6: Find the union:
\(x=\left\{ \begin{array}{l}-\frac{\arccos\left(-0.123\right)}{3}+\frac{2k\pi }{3}\\\frac{\arccos\left(-0.123\right)}{3}+\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\)
- step7: Simplify:
\(x\approx \left\{ \begin{array}{l}-0.564703+\frac{2k\pi }{3}\\0.564703+\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( 3\cos(x-15^{\text{°}})+1=-0.456 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(3\cos\left(x-15\right)+1=-0.456\)
- step1: Move the constant to the right side:
\(3\cos\left(x-15\right)=-0.456-1\)
- step2: Subtract the numbers:
\(3\cos\left(x-15\right)=-1.456\)
- step3: Divide both sides:
\(\frac{3\cos\left(x-15\right)}{3}=\frac{-1.456}{3}\)
- step4: Divide the numbers:
\(\cos\left(x-15\right)=-\frac{182}{375}\)
- step5: Use the inverse trigonometric function:
\(x-15=\arccos\left(-\frac{182}{375}\right)\)
- step6: Calculate:
\(\begin{align}&x-15=-\arccos\left(-\frac{182}{375}\right)\\&x-15=\arccos\left(-\frac{182}{375}\right)\end{align}\)
- step7: Add the period:
\(\begin{align}&x-15=-\arccos\left(-\frac{182}{375}\right)+2k\pi ,k \in \mathbb{Z}\\&x-15=\arccos\left(-\frac{182}{375}\right)+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step8: Calculate:
\(\begin{align}&x=-\arccos\left(-\frac{182}{375}\right)+15+2k\pi ,k \in \mathbb{Z}\\&x-15=\arccos\left(-\frac{182}{375}\right)+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step9: Calculate:
\(\begin{align}&x=-\arccos\left(-\frac{182}{375}\right)+15+2k\pi ,k \in \mathbb{Z}\\&x=\arccos\left(-\frac{182}{375}\right)+15+2k\pi ,k \in \mathbb{Z}\end{align}\)
- step10: Find the union:
\(x=\left\{ \begin{array}{l}-\arccos\left(-\frac{182}{375}\right)+15+2k\pi \\\arccos\left(-\frac{182}{375}\right)+15+2k\pi \end{array}\right.,k \in \mathbb{Z}\)
Here are the general solutions for the given equations:
1. **Equation:** \( \sin(x - 16^{\circ}) = 0.616 \)
**Solution:**
\[
x \approx \left\{
\begin{array}{l}
-0.330808 + k\pi \\
0.330808 + k\pi
\end{array}
\right., k \in \mathbb{Z}
\]
2. **Equation:** \( \cos(2x) = 0.789 \)
**Solution:**
\[
x \approx \left\{
\begin{array}{l}
-0.564703 + \frac{2k\pi}{3} \\
0.564703 + \frac{2k\pi}{3}
\end{array}
\right., k \in \mathbb{Z}
\]
3. **Equation:** \( \cos(3x) = -0.123 \)
**Solution:**
\[
x \approx \left\{
\begin{array}{l}
-0.564703 + \frac{2k\pi}{3} \\
0.564703 + \frac{2k\pi}{3}
\end{array}
\right., k \in \mathbb{Z}
\]
4. **Equation:** \( \tan(x + 56^{\circ}) = 7.56 \)
**Solution:**
\[
x \approx -54.560715 + k\pi, k \in \mathbb{Z}
\]
5. **Equation:** \( \tan\left(\frac{x}{2}\right) = -1.421 \)
**Solution:**
\[
x \approx -1.915143 + 2k\pi, k \in \mathbb{Z}
\]
6. **Equation:** \( \sin(2x + 44^{\circ}) = -0.708 \)
**Solution:**
\[
x \approx \left\{
\begin{array}{l}
-22.393331 + k\pi \\
-20.035873 + k\pi
\end{array}
\right., k \in \mathbb{Z}
\]
7. **Equation:** \( 3\cos(x - 15^{\circ}) + 1 = -0.456 \)
**Solution:**
\[
x = \left\{
\begin{array}{l}
-\arccos\left(-\frac{182}{375}\right) + 15 + 2k\pi \\
\arccos\left(-\frac{182}{375}\right) + 15 + 2k\pi
\end{array}
\right., k \in \mathbb{Z}
\]
8. **Equation:** \( 2\tan(2x - 10^{\circ}) = 10.67 \)
**Solution:**
\[
x = \frac{\arctan\left(\frac{1067}{200}\right) + 10}{2} + \frac{k\pi}{2}, k \in \mathbb{Z}
\]
These solutions provide the general forms for each equation, where \( k \) is any integer.
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