Claswork 10 Maxinzozs Determine the general solution a \( \sin x+2 \cos x=0 \) b, \( \cos (x+20)=-0,242 \) a \( \cos 2 x+1=0 \) d) \( \cos 4 x \cos x+\sin x(\sin 4 x=-07 \) e, \( \sin (2 x+40) \cos (x+30)-\cos (2 x+40) \sin (x+30)=\cos (2 x-20) \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(x+20\right)=-0.242\) - step1: Use the inverse trigonometric function: \(x+20=\arccos\left(-0.242\right)\) - step2: Calculate: \(\begin{align}&x+20=-\arccos\left(-0.242\right)\\&x+20=\arccos\left(-0.242\right)\end{align}\) - step3: Add the period: \(\begin{align}&x+20=-\arccos\left(-0.242\right)+2k\pi ,k \in \mathbb{Z}\\&x+20=\arccos\left(-0.242\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step4: Calculate: \(\begin{align}&x=-\arccos\left(-0.242\right)-20+2k\pi ,k \in \mathbb{Z}\\&x+20=\arccos\left(-0.242\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step5: Calculate: \(\begin{align}&x=-\arccos\left(-0.242\right)-20+2k\pi ,k \in \mathbb{Z}\\&x=\arccos\left(-0.242\right)-20+2k\pi ,k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}-\arccos\left(-0.242\right)-20+2k\pi \\\arccos\left(-0.242\right)-20+2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step7: Simplify: \(x\approx \left\{ \begin{array}{l}-21.815223+2k\pi \\-18.184777+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \sin x + 2 \cos x = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sin\left(x\right)+2\cos\left(x\right)=0\) - step1: Move the expression to the right side: \(2\cos\left(x\right)=0-\sin\left(x\right)\) - step2: Subtract the terms: \(2\cos\left(x\right)=-\sin\left(x\right)\) - step3: Divide both sides: \(\frac{2\cos\left(x\right)}{\sin\left(x\right)}=-1\) - step4: Divide the terms: \(2\cot\left(x\right)=-1\) - step5: Multiply both sides: \(2\cot\left(x\right)\times \frac{1}{2}=-\frac{1}{2}\) - step6: Calculate: \(\cot\left(x\right)=-\frac{1}{2}\) - step7: Use the inverse trigonometric function: \(x=\operatorname{arccot}\left(-\frac{1}{2}\right)\) - step8: Add the period: \(x=\operatorname{arccot}\left(-\frac{1}{2}\right)+k\pi ,k \in \mathbb{Z}\) Solve the equation \( \cos 2x + 1 = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(2x\right)+1=0\) - step1: Move the constant to the right side: \(\cos\left(2x\right)=0-1\) - step2: Remove 0: \(\cos\left(2x\right)=-1\) - step3: Use the inverse trigonometric function: \(2x=\arccos\left(-1\right)\) - step4: Calculate: \(2x=\pi \) - step5: Add the period: \(2x=\pi +2k\pi ,k \in \mathbb{Z}\) - step6: Solve the equation: \(x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) Solve the equation \( \cos 4x \cos x + \sin x (\sin 4x) = -0.7 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(4x\right)\cos\left(x\right)+\sin\left(x\right)\sin\left(4x\right)=-0.7\) - step1: Simplify: \(\cos\left(3x\right)=-0.7\) - step2: Use the inverse trigonometric function: \(3x=\arccos\left(-0.7\right)\) - step3: Calculate: \(\begin{align}&3x=-\arccos\left(-0.7\right)\\&3x=\arccos\left(-0.7\right)\end{align}\) - step4: Add the period: \(\begin{align}&3x=-\arccos\left(-0.7\right)+2k\pi ,k \in \mathbb{Z}\\&3x=\arccos\left(-0.7\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step5: Calculate: \(\begin{align}&x=-\frac{\arccos\left(-0.7\right)}{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\\&3x=\arccos\left(-0.7\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step6: Calculate: \(\begin{align}&x=-\frac{\arccos\left(-0.7\right)}{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\\&x=\frac{\arccos\left(-0.7\right)}{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\end{align}\) - step7: Find the union: \(x=\left\{ \begin{array}{l}-\frac{\arccos\left(-0.7\right)}{3}+\frac{2k\pi }{3}\\\frac{\arccos\left(-0.7\right)}{3}+\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\) - step8: Simplify: \(x\approx \left\{ \begin{array}{l}-0.782065+\frac{2k\pi }{3}\\0.782065+\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \sin (2x + 40) \cos (x + 30) - \cos (2x + 40) \sin (x + 30) = \cos (2x - 20) \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sin\left(2x+40\right)\cos\left(x+30\right)-\cos\left(2x+40\right)\sin\left(x+30\right)=\cos\left(2x-20\right)\) - step1: Simplify: \(\sin\left(x+10\right)=\cos\left(2x-20\right)\) - step2: Move the expression to the left side: \(\sin\left(x+10\right)-\cos\left(2x-20\right)=0\) - step3: Transform the expression: \(-2\sin\left(\frac{\pi -60+2x}{4}\right)\sin\left(\frac{\pi +20-6x}{4}\right)=0\) - step4: Elimination the left coefficient: \(\sin\left(\frac{\pi -60+2x}{4}\right)\sin\left(\frac{\pi +20-6x}{4}\right)=0\) - step5: Separate into possible cases: \(\begin{align}&\sin\left(\frac{\pi -60+2x}{4}\right)=0\\&\sin\left(\frac{\pi +20-6x}{4}\right)=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=\frac{-\pi +60}{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{\pi +20}{6}+\frac{2k\pi }{3},k \in \mathbb{Z}\end{align}\) - step7: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi +20}{6}+\frac{2k\pi }{3}\\\frac{-\pi +60}{2}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Here are the general solutions for each of the equations provided: ### a) \( \sin x + 2 \cos x = 0 \) To solve this equation, we can rearrange it as follows: \[ \sin x = -2 \cos x \] This can be expressed in terms of the cotangent: \[ \frac{\sin x}{\cos x} = -2 \implies \tan x = -\frac{1}{2} \] Thus, the general solution is: \[ x = \operatorname{arccot}(-\frac{1}{2}) + k\pi, \quad k \in \mathbb{Z} \] ### b) \( \cos (x + 20) = -0.242 \) To solve for \( x \), we take the inverse cosine: \[ x + 20 = \cos^{-1}(-0.242) \] Calculating the values gives: \[ x \approx \left\{ \begin{array}{l} -21.815223 + 2k\pi \\ -18.184777 + 2k\pi \end{array} \right., \quad k \in \mathbb{Z} \] ### c) \( \cos 2x + 1 = 0 \) Rearranging gives: \[ \cos 2x = -1 \] The general solution for this is: \[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \] ### d) \( \cos 4x \cos x + \sin x (\sin 4x) = -0.7 \) This equation can be solved numerically or graphically. The approximate solutions are: \[ x \approx \left\{ \begin{array}{l} -0.782065 + \frac{2k\pi}{3} \\ 0.782065 + \frac{2k\pi}{3} \end{array} \right., \quad k \in \mathbb{Z} \] ### e) \( \sin (2x + 40) \cos (x + 30) - \cos (2x + 40) \sin (x + 30) = \cos (2x - 20) \) This can be simplified using the sine subtraction formula: \[ \sin((2x + 40) - (x + 30)) = \cos(2x - 20) \] This leads to: \[ \sin(x + 10) = \cos(2x - 20) \] The general solutions for this equation are: \[ x = \left\{ \begin{array}{l} \frac{\pi + 20}{6} + \frac{2k\pi}{3} \\ \frac{-\pi + 60}{2} + 2k\pi \end{array} \right., \quad k \in \mathbb{Z} \] These solutions provide the general forms for each of the equations given.