$$ 1.2 \sin \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}(\sin x-\cos x) $$ $$ 1.3 \cdot \cos (x+y) \cos (x-y)=(\cos x \cos y)^{2}-(\sin x \sin y)^{2} $$ $$ 1.4 \frac{\cot x-\tan x}{\cot ^{2} x-\tan ^{2} x}=\frac{1}{2} \sin 2 x $$ 4 QuESTIOW 2 2. Prove that $$ \sin 61^{\circ}+\sin 29^{\circ}=\sqrt{2} \cos 16^{\circ} $$. QuたSTHON3 3. If $$ \tan x=-\frac{1}{2} $$ and $$ \tan y=-\frac{1}{3} $$ with both $$ x $$ and $$ y $$ in the second quadrant, calculate the value of $$ \cos (x+y) $$. A calculator may not be used. $$ \text { QuझSTIow } 4 $$ Solve for $$ x $$ if: $$ \operatorname{cosec} x+\cot x=\sqrt{3} ; 0^{\circ} \leq x \leq 360^{\circ} $$ (6) (4)

Question

UpStudy AI · Accepted Answer

Solve the system of equations by following steps:
- step0: Solution:
  $$\left\{ \begin{array}{l}\tan\left(x\right)=-\frac{1}{2}\\tan\left(y\right)=-\frac{1}{3}\end{array}\right.$$
- step1: Find the domain:
  $$\left\{ \begin{array}{l}\tan\left(x\right)=-\frac{1}{2}\\tan\left(y\right)=-\frac{1}{3}\end{array}\right.,x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\cap y\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step2: Calculate:
  $$\left\{ \begin{array}{l}x=\arctan\left(-\frac{1}{2}\right)+k\pi ,k \in \mathbb{Z}\y=\arctan\left(-\frac{1}{3}\right)+k\pi ,k \in \mathbb{Z}\end{array}\right.$$
- step3: Check if the solution is in the defined range:
  $$\left\{ \begin{array}{l}x=\arctan\left(-\frac{1}{2}\right)+k\pi ,k \in \mathbb{Z}\y=\arctan\left(-\frac{1}{3}\right)+k\pi ,k \in \mathbb{Z}\end{array}\right.,x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\cap y\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step4: Find the intersection:
  $$\left\{ \begin{array}{l}x=\arctan\left(-\frac{1}{2}\right)+k\pi ,k \in \mathbb{Z}\y=\arctan\left(-\frac{1}{3}\right)+k\pi ,k \in \mathbb{Z}\end{array}\right.$$
Solve the equation $$ \operatorname{cosec} x+\cot x=\sqrt{3} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\csc\left(x\right)+\cot\left(x\right)=\sqrt{3}$$
- step1: Find the domain:
  $$\csc\left(x\right)+\cot\left(x\right)=\sqrt{3},x\neq k\pi ,k \in \mathbb{Z}$$
- step2: Rewrite the expression:
  $$\frac{1}{\sin\left(x\right)}+\frac{\cos\left(x\right)}{\sin\left(x\right)}=\sqrt{3}$$
- step3: Multiply both sides of the equation by LCD:
  $$\left(\frac{1}{\sin\left(x\right)}+\frac{\cos\left(x\right)}{\sin\left(x\right)}\right)\sin\left(x\right)=\sqrt{3}\times \sin\left(x\right)$$
- step4: Simplify the equation:
  $$1+\cos\left(x\right)=\sqrt{3}\times \sin\left(x\right)$$
- step5: Raise both sides to the $$2$$-th power$$:$$
  $$\left(1+\cos\left(x\right)\right)^{2}=\left(\sqrt{3}\times \sin\left(x\right)\right)^{2}$$
- step6: Calculate:
  $$1+2\cos\left(x\right)+\cos^{2}\left(x\right)=3\sin^{2}\left(x\right)$$
- step7: Rewrite the expression:
  $$1+2\cos\left(x\right)+\cos^{2}\left(x\right)=3-3\cos^{2}\left(x\right)$$
- step8: Move the expression to the left side:
  $$1+2\cos\left(x\right)+\cos^{2}\left(x\right)-\left(3-3\cos^{2}\left(x\right)\right)=0$$
- step9: Calculate:
  $$-2+2\cos\left(x\right)+4\cos^{2}\left(x\right)=0$$
- step10: Factor the expression:
  $$2\left(-1-\cos\left(x\right)\right)\left(1-2\cos\left(x\right)\right)=0$$
- step11: Calculate:
  $$\left(-1-\cos\left(x\right)\right)\left(1-2\cos\left(x\right)\right)=0$$
- step12: Separate into possible cases:
  $$\begin{align}&-1-\cos\left(x\right)=0\&1-2\cos\left(x\right)=0\end{align}$$
- step13: Solve the equation:
  $$\begin{align}&x=\pi +2k\pi ,k \in \mathbb{Z}\&x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step14: Find the union:
  $$x=\frac{\pi }{3}+\frac{2k\pi }{3},k \in \mathbb{Z}$$
- step15: Check if the solution is in the defined range:
  $$x=\frac{\pi }{3}+\frac{2k\pi }{3},k \in \mathbb{Z},x\neq k\pi ,k \in \mathbb{Z}$$
- step16: Find the intersection:
  $$x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ \frac{\cot x-\tan x}{\cot ^{2} x-\tan ^{2} x}=\frac{1}{2} \sin 2 x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{\cot\left(x\right)-\tan\left(x\right)}{\cot^{2}\left(x\right)-\tan^{2}\left(x\right)}=\frac{1}{2}\sin\left(2x\right)$$
- step1: Find the domain:
  $$\frac{\cot\left(x\right)-\tan\left(x\right)}{\cot^{2}\left(x\right)-\tan^{2}\left(x\right)}=\frac{1}{2}\sin\left(2x\right),x\neq \frac{k\pi }{4},k \in \mathbb{Z}$$
- step2: Divide the terms:
  $$\frac{1}{\cot\left(x\right)+\tan\left(x\right)}=\frac{1}{2}\sin\left(2x\right)$$
- step3: Rewrite the expression:
  $$\frac{\sin\left(x\right)\cos\left(x\right)}{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}=\frac{1}{2}\sin\left(2x\right)$$
- step4: Rewrite the expression:
  $$\frac{\sin\left(x\right)\cos\left(x\right)}{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}=\frac{1}{2}\times 2\sin\left(x\right)\cos\left(x\right)$$
- step5: Simplify:
  $$\frac{\sin\left(x\right)\cos\left(x\right)}{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}=\sin\left(x\right)\cos\left(x\right)$$
- step6: Cross multiply:
  $$\sin\left(x\right)\cos\left(x\right)=\left(\cos^{2}\left(x\right)+\sin^{2}\left(x\right)\right)\sin\left(x\right)\cos\left(x\right)$$
- step7: Move the expression to the left side:
  $$\sin\left(x\right)\cos\left(x\right)-\left(\cos^{2}\left(x\right)+\sin^{2}\left(x\right)\right)\sin\left(x\right)\cos\left(x\right)=0$$
- step8: Calculate:
  $$\sin\left(x\right)\cos\left(x\right)+\left(-\cos^{2}\left(x\right)-\sin^{2}\left(x\right)\right)\sin\left(x\right)\cos\left(x\right)=0$$
- step9: Factor the expression:
  $$\sin\left(x\right)\cos\left(x\right)\left(1-\cos^{2}\left(x\right)-\sin^{2}\left(x\right)\right)=0$$
- step10: Separate into possible cases:
  $$\begin{align}&\sin\left(x\right)=0\&\cos\left(x\right)=0\&1-\cos^{2}\left(x\right)-\sin^{2}\left(x\right)=0\end{align}$$
- step11: Solve the equation:
  $$\begin{align}&x=k\pi ,k \in \mathbb{Z}\&x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\&x \in \mathbb{R}\end{align}$$
- step12: Find the union:
  $$x \in \mathbb{R}$$
- step13: Check if the solution is in the defined range:
  $$x \in \mathbb{R},x\neq \frac{k\pi }{4},k \in \mathbb{Z}$$
- step14: Find the intersection:
  $$x\neq \frac{k\pi }{4},k \in \mathbb{Z}$$
Solve the equation $$ 1.2 \sin \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}(\sin x-\cos x) $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$1.2\sin\left(x-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\left(\sin\left(x\right)-\cos\left(x\right)\right)$$
- step1: Use the periodicity identities:
  $$1.2\sin\left(x+\frac{7\pi }{4}\right)=\frac{\sqrt{2}}{2}\sin\left(x\right)-\frac{\sqrt{2}}{2}\cos\left(x\right)$$
- step2: Move the expression to the left side:
  $$1.2\sin\left(x+\frac{7\pi }{4}\right)-\left(\frac{\sqrt{2}}{2}\sin\left(x\right)-\frac{\sqrt{2}}{2}\cos\left(x\right)\right)=0$$
- step3: Calculate:
  $$1.2\sin\left(x+\frac{7\pi }{4}\right)-\frac{\sqrt{2}}{2}\sin\left(x\right)+\frac{\sqrt{2}}{2}\cos\left(x\right)=0$$
- step4: Simplify:
  $$\frac{\sqrt{2}}{10}\sin\left(x\right)-\frac{\sqrt{2}}{10}\cos\left(x\right)=0$$
- step5: Solve using substitution:
  $$\frac{\sqrt{2}}{10}\times \frac{2\tan\left(\frac{1}{2}x\right)}{1+\tan^{2}\left(\frac{1}{2}x\right)}-\frac{\sqrt{2}}{10}\times \frac{1-\tan^{2}\left(\frac{1}{2}x\right)}{1+\tan^{2}\left(\frac{1}{2}x\right)}=0$$
- step6: Solve using substitution:
  $$\frac{\sqrt{2}}{10}\times \frac{2t}{1+t^{2}}-\frac{\sqrt{2}}{10}\times \frac{1-t^{2}}{1+t^{2}}=0$$
- step7: Rewrite the expression:
  $$\frac{2\sqrt{2}\times t}{10\left(1+t^{2}\right)}-\frac{\sqrt{2}\times \left(1-t^{2}\right)}{10\left(1+t^{2}\right)}=0$$
- step8: Multiply both sides of the equation by LCD:
  $$\left(\frac{2\sqrt{2}\times t}{10\left(1+t^{2}\right)}-\frac{\sqrt{2}\times \left(1-t^{2}\right)}{10\left(1+t^{2}\right)}\right)\times 10\left(1+t^{2}\right)=0\times 10\left(1+t^{2}\right)$$
- step9: Simplify the equation:
  $$2\sqrt{2}\times t-\sqrt{2}+\sqrt{2}\times t^{2}=0$$
- step10: Rewrite in standard form:
  $$\sqrt{2}\times t^{2}+2\sqrt{2}\times t-\sqrt{2}=0$$
- step11: Solve using the quadratic formula:
  $$t=\frac{-2\sqrt{2}\pm \sqrt{\left(2\sqrt{2}\right)^{2}-4\sqrt{2}\times \left(-\sqrt{2}\right)}}{2\sqrt{2}}$$
- step12: Simplify the expression:
  $$t=\frac{-2\sqrt{2}\pm \sqrt{16}}{2\sqrt{2}}$$
- step13: Simplify the expression:
  $$t=\frac{-2\sqrt{2}\pm 4}{2\sqrt{2}}$$
- step14: Separate into possible cases:
  $$\begin{align}&t=\frac{-2\sqrt{2}+4}{2\sqrt{2}}\&t=\frac{-2\sqrt{2}-4}{2\sqrt{2}}\end{align}$$
- step15: Simplify the expression:
  $$\begin{align}&t=-1+\sqrt{2}\&t=\frac{-2\sqrt{2}-4}{2\sqrt{2}}\end{align}$$
- step16: Simplify the expression:
  $$\begin{align}&t=-1+\sqrt{2}\&t=-1-\sqrt{2}\end{align}$$
- step17: Substitute back:
  $$\begin{align}&\tan\left(\frac{1}{2}x\right)=-1+\sqrt{2}\&\tan\left(\frac{1}{2}x\right)=-1-\sqrt{2}\end{align}$$
- step18: Calculate:
  $$\begin{align}&x=\frac{\pi }{4}+2k\pi ,k \in \mathbb{Z}\&\tan\left(\frac{1}{2}x\right)=-1-\sqrt{2}\end{align}$$
- step19: Calculate:
  $$\begin{align}&x=\frac{\pi }{4}+2k\pi ,k \in \mathbb{Z}\&x=\frac{5\pi }{4}+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step20: Check if $$x=\pi+2k\pi,k\in\mathbb{Z}$$ is a solution$$:$$
  $$1.2\sin\left(\pi +2k\pi-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\left(\sin\left(\pi +2k\pi\right)-\cos\left(\pi +2k\pi\right)\right)$$
- step21: Calculate:
  $$1.2\sin\left(\pi -\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\left(\sin\left(\pi \right)-\cos\left(\pi \right)\right)$$
- step22: Simplify:
  $$\frac{3\sqrt{2}}{5}=\frac{\sqrt{2}}{2}$$
- step23: Calculate:
  $$0.848528=\frac{\sqrt{2}}{2}$$
- step24: Calculate:
  $$0.848528=0.707107$$
- step25: Check the equality:
  $$\textrm{false}$$
- step26: Since $$x=\pi+2k\pi,k\in\mathbb{Z}$$ is not a solution,don't include it$$:$$
  $$\begin{align}&x=\frac{\pi }{4}+2k\pi ,k \in \mathbb{Z}\&x=\frac{5\pi }{4}+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step27: Find the union:
  $$x=\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}$$
Let's solve each part of the problem step by step.

### Question 1
1. **Equation**: 
   $$
   1.2 \sin \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}(\sin x-\cos x)
   $$
   **Solution**: 
   The solution gives:
   $$
   x=\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}
   $$

2. **Equation**: 
   $$
   1.3 \cdot \cos (x+y) \cos (x-y)=(\cos x \cos y)^{2}-(\sin x \sin y)^{2}
   $$
   **Solution**: 
   This equation timed out during calculation, indicating it may be too complex to solve directly. Further simplification or a different approach may be needed.

3. **Equation**: 
   $$
   \frac{\cot x-\tan x}{\cot ^{2} x-\tan ^{2} x}=\frac{1}{2} \sin 2 x
   $$
   **Solution**: 
   The result indicates:
   $$
   x\neq \frac{k\pi }{4},k \in \mathbb{Z}
   $$

4. **Equation**: 
   $$
   \sin 61^{\circ}+\sin 29^{\circ}=\sqrt{2} \cos 16^{\circ}
   $$
   **Solution**: 
   This equation requires further analysis as it timed out during calculation.

### Question 2
5. **Given**: 
   $$
   \tan x=-\frac{1}{2}, \tan y=-\frac{1}{3}
   $$
   **Both $$x$$ and $$y$$ are in the second quadrant**. 
   **Solution**: 
   The solutions for $$x$$ and $$y$$ are:
   $$
   \left\{ 
   \begin{array}{l}
   x=\arctan(-\frac{1}{2})+k\pi ,k \in \mathbb{Z} \
   y=\arctan(-\frac{1}{3})+k\pi ,k \in \mathbb{Z}
   \end{array}
   \right.
   $$

### Question 3
6. **Equation**: 
   $$
   \operatorname{cosec} x+\cot x=\sqrt{3} ; 0^{\circ} \leq x \leq 360^{\circ}
   $$
   **Solution**: 
   The solutions for this equation are:
   $$
   x=\left\{ 
   \begin{array}{l}
   \frac{\pi }{3}+2k\pi \
   \frac{5\pi }{3}+2k\pi 
   \end{array}
   \right., k \in \mathbb{Z}
   $$

### Summary of Results
- For the first equation, $$ x=\frac{\pi }{4}+k\pi $$.
- The second equation requires further analysis.
- The third equation indicates $$ x\neq \frac{k\pi }{4} $$.
- The fourth equation requires further analysis.
- For the tangent values, $$ x $$ and $$ y $$ are defined in terms of arctangent.
- The cosecant equation has solutions at specific angles.

If you need further assistance with any specific part or additional calculations, please let me know!
Here are the simplified solutions: 1. **First Equation**: $$ x = \frac{\pi}{4} + k\pi \quad \text{for any integer } k $$ 2. **Second Equation**: Requires further analysis as it timed out during calculation. 3. **Third Equation**: $$ x \neq \frac{k\pi}{4} \quad \text{for any integer } k $$ 4. **Fourth Equation**: Requires further analysis as it timed out during calculation. 5. **Second Question**: $$ x = \arctan\left(-\frac{1}{2}\right) + k\pi \quad \text{and} \quad y = \arctan\left(-\frac{1}{3}\right) + k\pi \quad \text{for any integer } k $$ 6. **Third Question**: $$ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{3} + 2k\pi \quad \text{for any integer } k $$ If you need more detailed steps or assistance with the equations that timed out, feel free to ask!

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