Pregunta

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

Ask by Wood Chadwick. in Kenya
Feb 26,2025

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a) \( \theta \) values: approximately \( 30^\circ \) and \( 150^\circ \) b) \( x \) values: approximately \( 60^\circ \) and \( 300^\circ \)

Solución

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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Mind Expander

To tackle the equation \( 2.5 \cos \theta - 4.2 \sin \theta = 3.6 \), we can rearrange it to isolate \(\theta\). Start by rewriting it as \( 2.5 \cos \theta = 4.2 \sin \theta + 3.6 \) and then express it in terms of tangent or use a substitution like \( R \cos(\theta + \phi) = k \), finding an appropriate \( R \) and \( \phi \) to solve for \(\theta\). This may involve some trigonometric identities and can lead you to possible angles for \(\theta\). For the second part \( 12 \sec x - 1 = 7 \tan^2 x \), remember that \(\sec x = \frac{1}{\cos x}\) and \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\). Start by substituting \( \sec^2 x - 1 = \tan^2 x \) to manipulate the equation. Rearranging the equation and solving for specific values leads to solutions within the interval: checking for all angles through substitution and identities ensures complete and accurate calculations. Have fun working through these angle transformations, and remember that practice will solidify your understanding and skills!

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