Acceso
Prima
Hogar Banco de preguntas Matemáticas Trigonometría
Pregunta

\( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \)

Ask by Martinez Bush. in South Africa
Mar 11,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

The solution to the equation is all angles \( \alpha \) except \( \alpha = \frac{3\pi}{2} + 2k\pi \) for any integer \( k \).

Solución

Solve the equation \( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \). Solve the equation by following steps: - step0: Solve for \(\alpha\): \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\frac{2\sin\left(\alpha \right)-1}{\sin\left(\alpha \right)-2}\) - step1: Find the domain: \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\frac{2\sin\left(\alpha \right)-1}{\sin\left(\alpha \right)-2},\alpha \neq \frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\left(\cos\left(2\alpha \right)-\sin\left(\alpha \right)\right)\left(1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)\right)^{-1}\) - step3: Express with a positive exponent: \(\left(\cos\left(2\alpha \right)-\sin\left(\alpha \right)\right)\times \frac{1}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}\) - step4: Rewrite the expression: \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}\) - step5: Rewrite the expression: \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\left(2\sin\left(\alpha \right)-1\right)\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step6: Rewrite the expression: \(\frac{2\cos^{2}\left(\alpha \right)-1-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\left(2\sin\left(\alpha \right)-1\right)\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step7: Expand the expression: \(\frac{2\cos^{2}\left(\alpha \right)-1-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=2\sin\left(\alpha \right)\left(\sin\left(\alpha \right)-2\right)^{-1}-\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step8: Rewrite the expression: \(\frac{1-2\sin\left(\alpha \right)}{2-\sin\left(\alpha \right)}=2\sin\left(\alpha \right)\left(\sin\left(\alpha \right)-2\right)^{-1}-\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step9: Rewrite the expression: \(\frac{1-2\sin\left(\alpha \right)}{2-\sin\left(\alpha \right)}=\frac{2\sin\left(\alpha \right)}{\sin\left(\alpha \right)-2}-\frac{1}{\sin\left(\alpha \right)-2}\) - step10: Multiply both sides of the equation by LCD: \(\frac{1-2\sin\left(\alpha \right)}{2-\sin\left(\alpha \right)}\times \left(-\sin\left(\alpha \right)+2\right)=\left(\frac{2\sin\left(\alpha \right)}{\sin\left(\alpha \right)-2}-\frac{1}{\sin\left(\alpha \right)-2}\right)\left(-\sin\left(\alpha \right)+2\right)\) - step11: Simplify the equation: \(1-2\sin\left(\alpha \right)=-2\sin\left(\alpha \right)+1\) - step12: Move the expression to the left side: \(1-2\sin\left(\alpha \right)-\left(-2\sin\left(\alpha \right)+1\right)=0\) - step13: Calculate: \(0=0\) - step14: The statement is true: \(\alpha \in \mathbb{R}\) - step15: Check if the solution is in the defined range: \(\alpha \in \mathbb{R},\alpha \neq \frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\) - step16: Find the intersection: \(\alpha \neq \frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\) The solution to the equation \( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \) is \( \alpha \neq \frac{3\pi}{2}+2k\pi \) where \( k \in \mathbb{Z} \).

Respondido por UpStudy AI y revisado por un tutor profesional

error msg
Explicar
Simplifique esta solución

Mind Expander

To solve the equation \( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \), we can begin with some trigonometric identities and transformations. First, recall that \( \cos 2\alpha = 1 - 2\sin^2\alpha \). So we can substitute this into the equation: \[ \frac{(1 - 2\sin^2 \alpha) - \sin \alpha}{1+\sin \alpha + \cos^2 \alpha} \] Now, observe that \( \cos^2 \alpha = 1 - \sin^2 \alpha \), hence \( 1 + \sin \alpha + \cos^2 \alpha = 1 + \sin \alpha + (1 - \sin^2 \alpha) = 2 + \sin \alpha - \sin^2 \alpha \). Thus, the left side becomes: \[ \frac{1 - 2\sin^2 \alpha - \sin \alpha}{2 + \sin \alpha - \sin^2 \alpha} \] We can now work on simplifying both sides. For the right side, notice that if we rearrange, we will have: \[ \frac{2\sin\alpha - 1}{\sin\alpha - 2} \] Next, we can cross-multiply to eliminate the fractions: \[ (1 - 2\sin^2 \alpha - \sin \alpha)(\sin \alpha - 2) = (2\sin \alpha - 1)(2 + \sin \alpha - \sin^2 \alpha) \] Expanding both sides will yield a polynomial equation where you can collect all terms to one side, simplify, and then solve for \(\alpha\). However, a quicker numerical or graphical approach can also be used if feasible, by plotting both sides of the original equation to find potential solutions for \( \alpha \). Have fun solving it, and remember to check for extraneous solutions that might arise when squaring or simplifying!

preguntas relacionadas

5) \( 45^{\circ} \times \frac{\pi}{180^{\circ}}= \)
Trigonometría Colombia Mar 19, 2025
c) \( \cos 240^{\circ}+\sin 210^{\circ}+\frac{\sin 310^{\circ}}{\cos 140^{\circ}} \) d) \( \frac{\sin ^{2}\left(-100^{\circ}\right)}{\cos 120^{\circ} \sin 80^{\circ} \cos 370^{\circ}} \) e) \( \frac{\sin \left(-640^{\circ}\right)}{\sin ^{2} 217^{\circ}-\cos \left(-397^{\circ}\right) \sin 307^{\circ}} \times \frac{1}{\sin 260^{\circ}} \) 12. Simplify: a) \( \left[\sin (-\theta)+\cos \left(360^{\circ}+\theta\right)\right]\left[\cos \left(\theta-90^{\circ}\right)+\cos \theta\right] \) b) \( \frac{\sin \left(90^{\circ}-x\right) \sin \left(90^{\circ}+x\right)}{\left(\sin 90^{\circ}-\sin x\right)\left(\sin 90^{\circ}+\sin x\right)} \) c) \( \frac{\cos 140^{\circ}-\sin \left(90^{\circ}-x\right)}{\sin 410^{\circ}+\cos (-x)} \) d) \( \cos \left(x-90^{\circ}\right) \sin \left(x-180^{\circ}\right)+\frac{\cos \left(360^{\circ}+x\right)}{\sin \left(90^{\circ}-x\right)} \) e) \( \frac{\cos \left(90^{\circ}+x\right) \cos (-x) \sin (-x)}{\sin \left(x-90^{\circ}\right) \tan \left(360^{\circ}-x\right) \cos x} \) f) \( 1-\frac{\sin ^{2}\left(180^{\circ}+x\right)}{1-\cos ^{2}\left(180^{\circ}+x\right)} \)
Trigonometría South Africa Mar 19, 2025
Prove that \( \frac{\operatorname{Sin} \theta(1+\operatorname{Cos} \theta)}{1-\cos ^{2} \theta}+\tan \theta=\frac{1+\operatorname{Cos}}{\sin \theta \cdot Q} \)
Trigonometría South Africa Mar 19, 2025
7. Найдите значение выражения \( 4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}+2,5 \) \( \begin{array}{llll}\text { A) } 3 & \text { B) } 3,5 & \text { C) } 3.75 & \text { D) } 4\end{array} \)
Trigonometría Azerbaijan Mar 19, 2025
Simplig the ff: friconometric expressin \( \frac{\cos \left(360^{\circ}-x\right) \cdot \operatorname{Sin}\left(360^{\circ}-x\right) \cdot \operatorname{Cos}\left(90^{\circ}-x\right)}{\sin ^{2}\left(00^{\circ}-x\right) \cdot \operatorname{Sin}\left(180^{\circ}+x\right)} \)
Trigonometría South Africa Mar 19, 2025

Latest Trigonometry Questions

7. Найдите значение выражения \( 4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}+2,5 \) \( \begin{array}{llll}\text { A) } 3 & \text { B) } 3,5 & \text { C) } 3.75 & \text { D) } 4\end{array} \)
Trigonometría Azerbaijan Mar 19, 2025
\( \arcsin \left(\cos \frac{3 \pi}{10}\right) \) \( \frac{\pi}{5} \) \( \frac{3 \pi}{10} \) \( \frac{31}{2} \) Найти наименьшее положительное решение уравнения * (Баллов: 1) \( \sin ^{2} x+1.5 \sin x-1=0 \) 0 \( \frac{\pi}{3} \) \( \frac{\pi}{6} \) \( \frac{\pi}{2} \)
Trigonometría Uzbekistan Mar 19, 2025
Prove that \( \frac{\operatorname{Sin} \theta(1+\operatorname{Cos} \theta)}{1-\cos ^{2} \theta}+\tan \theta=\frac{1+\operatorname{Cos}}{\sin \theta \cdot Q} \)
Trigonometría South Africa Mar 19, 2025
Simplig the ff: friconometric expressin \( \frac{\cos \left(360^{\circ}-x\right) \cdot \operatorname{Sin}\left(360^{\circ}-x\right) \cdot \operatorname{Cos}\left(90^{\circ}-x\right)}{\sin ^{2}\left(00^{\circ}-x\right) \cdot \operatorname{Sin}\left(180^{\circ}+x\right)} \)
Trigonometría South Africa Mar 19, 2025
c) \( \cos 240^{\circ}+\sin 210^{\circ}+\frac{\sin 310^{\circ}}{\cos 140^{\circ}} \) d) \( \frac{\sin ^{2}\left(-100^{\circ}\right)}{\cos 120^{\circ} \sin 80^{\circ} \cos 370^{\circ}} \) e) \( \frac{\sin \left(-640^{\circ}\right)}{\sin ^{2} 217^{\circ}-\cos \left(-397^{\circ}\right) \sin 307^{\circ}} \times \frac{1}{\sin 260^{\circ}} \) 12. Simplify: a) \( \left[\sin (-\theta)+\cos \left(360^{\circ}+\theta\right)\right]\left[\cos \left(\theta-90^{\circ}\right)+\cos \theta\right] \) b) \( \frac{\sin \left(90^{\circ}-x\right) \sin \left(90^{\circ}+x\right)}{\left(\sin 90^{\circ}-\sin x\right)\left(\sin 90^{\circ}+\sin x\right)} \) c) \( \frac{\cos 140^{\circ}-\sin \left(90^{\circ}-x\right)}{\sin 410^{\circ}+\cos (-x)} \) d) \( \cos \left(x-90^{\circ}\right) \sin \left(x-180^{\circ}\right)+\frac{\cos \left(360^{\circ}+x\right)}{\sin \left(90^{\circ}-x\right)} \) e) \( \frac{\cos \left(90^{\circ}+x\right) \cos (-x) \sin (-x)}{\sin \left(x-90^{\circ}\right) \tan \left(360^{\circ}-x\right) \cos x} \) f) \( 1-\frac{\sin ^{2}\left(180^{\circ}+x\right)}{1-\cos ^{2}\left(180^{\circ}+x\right)} \)
Trigonometría South Africa Mar 19, 2025
Anterior Próximo
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad