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9. Solve a) \( 2.5 \cos \theta-4.2 \sin \theta=3.6: 0^{\circ} \leq \theta \leq 360 \) b) \( 12 \operatorname{Sec} x-1=7 \tan ^{2} x: 0^{\circ} \leq \theta \leq 360^{\circ} \)

Ask by Savage Stewart. in Kenya
Feb 26,2025

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Answer

The solutions to the equations are: a) \( \theta = -\arcsin\left(\frac{1512 - \sqrt{1512^2 - 1603019}}{2389}\right) + 2k\pi \) and \( \theta = \arcsin\left(\frac{1512 + \sqrt{1512^2 - 1603019}}{2389}\right) + \pi + 2k\pi \), where \( k \) is an integer. b) \( x = -\operatorname{arcsec}\left(\frac{6 + \sqrt{78}}{7}\right) + 2k\pi \) and \( x = \operatorname{arcsec}\left(\frac{6 + \sqrt{78}}{7}\right) + 2k\pi \), where \( k \) is an integer.

Solution

Solve the equation \( 2.5\cos(\theta)-4.2\sin(\theta)=3.6 \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(2.5\cos\left(\theta \right)-4.2\sin\left(\theta \right)=3.6\) - step1: Move the expression to the right side: \(-4.2\sin\left(\theta \right)=3.6-2.5\cos\left(\theta \right)\) - step2: Move the expression to the left side: \(-4.2\sin\left(\theta \right)-3.6=-2.5\cos\left(\theta \right)\) - step3: Raise both sides to the \(2\)-th power\(:\) \(\left(-4.2\sin\left(\theta \right)-3.6\right)^{2}=\left(-2.5\cos\left(\theta \right)\right)^{2}\) - step4: Calculate: \(17.64\sin^{2}\left(\theta \right)+30.24\sin\left(\theta \right)+12.96=\frac{25}{4}\cos^{2}\left(\theta \right)\) - step5: Rewrite the expression: \(17.64\sin^{2}\left(\theta \right)+30.24\sin\left(\theta \right)+12.96=\frac{25}{4}-\frac{25}{4}\sin^{2}\left(\theta \right)\) - step6: Move the expression to the left side: \(17.64\sin^{2}\left(\theta \right)+30.24\sin\left(\theta \right)+12.96-\left(\frac{25}{4}-\frac{25}{4}\sin^{2}\left(\theta \right)\right)=0\) - step7: Calculate: \(\frac{2389}{100}\sin^{2}\left(\theta \right)+30.24\sin\left(\theta \right)+\frac{671}{100}=0\) - step8: Convert the decimal into a fraction: \(\frac{2389}{100}\sin^{2}\left(\theta \right)+\frac{756}{25}\sin\left(\theta \right)+\frac{671}{100}=0\) - step9: Multiply both sides: \(100\left(\frac{2389}{100}\sin^{2}\left(\theta \right)+\frac{756}{25}\sin\left(\theta \right)+\frac{671}{100}\right)=100\times 0\) - step10: Calculate: \(2389\sin^{2}\left(\theta \right)+3024\sin\left(\theta \right)+671=0\) - step11: Solve using the quadratic formula: \(\sin\left(\theta \right)=\frac{-3024\pm \sqrt{3024^{2}-4\times 2389\times 671}}{2\times 2389}\) - step12: Simplify the expression: \(\sin\left(\theta \right)=\frac{-3024\pm \sqrt{3024^{2}-4\times 2389\times 671}}{4778}\) - step13: Simplify the expression: \(\sin\left(\theta \right)=\frac{-3024\pm \sqrt{3024^{2}-6412076}}{4778}\) - step14: Simplify the expression: \(\sin\left(\theta \right)=\frac{-3024\pm 2\sqrt{1512^{2}-1603019}}{4778}\) - step15: Separate into possible cases: \(\begin{align}&\sin\left(\theta \right)=\frac{-3024+2\sqrt{1512^{2}-1603019}}{4778}\\&\sin\left(\theta \right)=\frac{-3024-2\sqrt{1512^{2}-1603019}}{4778}\end{align}\) - step16: Simplify the expression: \(\begin{align}&\sin\left(\theta \right)=\frac{-1512+\sqrt{1512^{2}-1603019}}{2389}\\&\sin\left(\theta \right)=\frac{-3024-2\sqrt{1512^{2}-1603019}}{4778}\end{align}\) - step17: Simplify the expression: \(\begin{align}&\sin\left(\theta \right)=\frac{-1512+\sqrt{1512^{2}-1603019}}{2389}\\&\sin\left(\theta \right)=-\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\end{align}\) - step18: Calculate: \(\begin{align}&\theta =\left\{ \begin{array}{l}-\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+2k\pi \\\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\\&\sin\left(\theta \right)=-\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\end{align}\) - step19: Calculate: \(\begin{align}&\theta =\left\{ \begin{array}{l}-\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+2k\pi \\\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\\&\theta =\left\{ \begin{array}{l}-\arcsin\left(\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\right)+2k\pi \\\arcsin\left(\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step20: Calculate: \(\theta =\left\{ \begin{array}{l}-\arcsin\left(\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\right)+2k\pi \\-\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+2k\pi \\\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+\pi +2k\pi \\\arcsin\left(\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step21: Check the solution: \(\theta =\left\{ \begin{array}{l}-\arcsin\left(\frac{1512-\sqrt{1512^{2}-1603019}}{2389}\right)+2k\pi \\\arcsin\left(\frac{1512+\sqrt{1512^{2}-1603019}}{2389}\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( 12\sec(x)-1=7\tan^{2}(x) \). Solve the equation by following steps: - step0: Solve for \(x\): \(12\sec\left(x\right)-1=7\tan^{2}\left(x\right)\) - step1: Find the domain: \(12\sec\left(x\right)-1=7\tan^{2}\left(x\right),x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step2: Rewrite the expression: \(12\sec\left(x\right)-1=7\sec^{2}\left(x\right)-7\) - step3: Move the expression to the left side: \(12\sec\left(x\right)-1-\left(7\sec^{2}\left(x\right)-7\right)=0\) - step4: Calculate: \(12\sec\left(x\right)+6-7\sec^{2}\left(x\right)=0\) - step5: Rewrite in standard form: \(-7\sec^{2}\left(x\right)+12\sec\left(x\right)+6=0\) - step6: Multiply both sides: \(7\sec^{2}\left(x\right)-12\sec\left(x\right)-6=0\) - step7: Solve using the quadratic formula: \(\sec\left(x\right)=\frac{12\pm \sqrt{\left(-12\right)^{2}-4\times 7\left(-6\right)}}{2\times 7}\) - step8: Simplify the expression: \(\sec\left(x\right)=\frac{12\pm \sqrt{\left(-12\right)^{2}-4\times 7\left(-6\right)}}{14}\) - step9: Simplify the expression: \(\sec\left(x\right)=\frac{12\pm \sqrt{312}}{14}\) - step10: Simplify the expression: \(\sec\left(x\right)=\frac{12\pm 2\sqrt{78}}{14}\) - step11: Separate into possible cases: \(\begin{align}&\sec\left(x\right)=\frac{12+2\sqrt{78}}{14}\\&\sec\left(x\right)=\frac{12-2\sqrt{78}}{14}\end{align}\) - step12: Simplify the expression: \(\begin{align}&\sec\left(x\right)=\frac{6+\sqrt{78}}{7}\\&\sec\left(x\right)=\frac{12-2\sqrt{78}}{14}\end{align}\) - step13: Simplify the expression: \(\begin{align}&\sec\left(x\right)=\frac{6+\sqrt{78}}{7}\\&\sec\left(x\right)=\frac{6-\sqrt{78}}{7}\end{align}\) - step14: Rearrange the terms: \(\begin{align}&\sec\left(x\right)=\frac{6+\sqrt{78}}{7}\\&x \notin \mathbb{R}\end{align}\) - step15: Calculate: \(\begin{align}&x=\left\{ \begin{array}{l}-\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \\\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\\&x \notin \mathbb{R}\end{align}\) - step16: Find the union: \(x=\left\{ \begin{array}{l}-\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \\\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step17: Check if the solution is in the defined range: \(x=\left\{ \begin{array}{l}-\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \\\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z},x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step18: Find the intersection: \(x=\left\{ \begin{array}{l}-\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \\\operatorname{arcsec}\left(\frac{6+\sqrt{78}}{7}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The solutions to the given trigonometric equations are: a) \( 2.5 \cos \theta - 4.2 \sin \theta = 3.6 \): \[ \theta = \left\{ \begin{array}{l} -\arcsin(\frac{1512-\sqrt{1512^{2}-1603019}}{2389})+2k\pi \\ \arcsin(\frac{1512+\sqrt{1512^{2}-1603019}}{2389})+\pi +2k\pi \end{array} \right., k \in \mathbb{Z} \] b) \( 12 \sec x - 1 = 7 \tan^{2} x \): \[ x = \left\{ \begin{array}{l} -\operatorname{arcsec}(\frac{6+\sqrt{78}}{7})+2k\pi \\ \operatorname{arcsec}(\frac{6+\sqrt{78}}{7})+2k\pi \end{array} \right., k \in \mathbb{Z} \]

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To solve the first equation \(2.5 \cos \theta - 4.2 \sin \theta = 3.6\), we can rearrange it to the form \(R \cos(\theta + \phi) = C\). To find \(R\), we calculate \(R = \sqrt{(2.5)^2 + (-4.2)^2} \approx 4.84\). Then, \(C = 3.6\) and we find \(\phi\) such that \(\tan \phi = \frac{-4.2}{2.5}\). Solving for \(\theta\), we can find the angles in the specified range where \( \theta = \phi + \cos^{-1}(\frac{C}{R})\) and \(\theta = 360^\circ - \phi - \cos^{-1}(\frac{C}{R})\). For the second equation \(12 \operatorname{Sec} x - 1 = 7 \tan^2 x\), we first rewrite \(\sec x\) as \(\frac{1}{\cos x}\) and \(\tan^2 x\) as \(\frac{\sin^2 x}{\cos^2 x}\). This gives us \(12 \cdot \frac{1}{\cos x} - 1 = 7 \cdot \frac{\sin^2 x}{\cos^2 x}\). By substituting \(\sin^2 x = 1 - \cos^2 x\), we can rearrange the equation into a polynomial inequality in terms of \(\cos x\). Solving this polynomial allows us to find the values of \(x\) in the required range. Don't forget to convert back to degrees or evaluate any other trigonometric expressions needed! Now, go grab your calculator and solve it step-by-step; the angles won’t find themselves!

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