Trigonometry. Graph y=cos (2x)-1. y = cos (2x) − 1 y = cos ( 2 x) - 1. Use the form acos(bx−c)+ d a cos ( b x - c) + d to find the variables used to find the amplitude, period, phase shift, and vertical shift. a = 1 a = 1. b = 2 b = 2. c = 0 c = 0. d = −1 d = - 1. Find the amplitude |a| | a |. Evaluate the integral. ∫ ( cos 2 x - 1) ( cos 2 x + 1) d x. = – ∫ ( 2 sin 2 x) ( 2 cos 2 x) d x = – ∫ tan 2 x d x = ∫ ( 1 – s e c 2 x) d x = x – tan x + C. Rewrite cos(2x) cos(x) cos ( 2 x) cos ( x) as a product. cos(2x) 1 cos(x) cos ( 2 x) 1 cos ( x) Write cos(2x) cos ( 2 x) as a fraction with denominator 1 1. cos(2x) 1 ⋅ 1 cos(x) cos ( 2 x) 1 ⋅ 1 cos ( x) Simplify. Tap for more steps cos(2x)sec(x) cos ( 2 x) sec ( x) Free math problem solver answers your algebra, geometry, trigonometry Answer link. The answer is =1+sinx We need a^2-b^2= (a+b) (a-b) We use cos^2x+sin^2x=1 cos^2x=1-sin^2x= (1+sinx) (1-sinx) Therefore, cos^2x/ (1-sinx)= ( (1+sinx)cancel (1-sinx))/cancel (1-sinx) =1+sinx. cos1(2x)cos(2x) cos 1 ( 2 x) cos ( 2 x) Raise cos(2x) cos ( 2 x) to the power of 1 1. cos1(2x)cos1(2x) cos 1 ( 2 x) cos 1 ( 2 x) Use the power rule aman = am+n a m a n = a m + n to combine exponents. cos(2x)1+1 cos ( 2 x) 1 + 1. Add 1 1 and 1 1. cos2(2x) cos 2 ( 2 x) Trigonometry. Solve for x 1-sin (x)=cos (2x) 1 − sin(x) = cos (2x) 1 - sin ( x) = cos ( 2 x) Subtract cos(2x) cos ( 2 x) from both sides of the equation. 1−sin(x)− cos(2x) = 0 1 - sin ( x) - cos ( 2 x) = 0. Simplify the left side of the equation. Tap for more steps −sin(x)+ 2sin2(x) = 0 - sin ( x) + 2 sin 2 ( x) = 0. . So for this question you can use either the product rule or the quotient rule and I'll run through them the quotient rule:The quotient rule says that if you have h(x)=f(x)/g(x)Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2So using f(x)=cos(2x) and g(x)=x^1/2then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2Plugging this into our formula gives ush(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xAlways remember to simplify afterwards which gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xSecond the product rule:What the product rule says is that ifh(x) = f(x)g(x)then h'(x) = f(x)g'(x) + f'(x)g(x)So if we say that h(x) = cos(2x)/x^1/2Then we can say that f(x) = cos(2x) and g(x) = x^-1/2Using the product rule we have:f(x) = cos(2x) f'(x) = -2sin(2x)g(x) = x^-1/2 g'(x) = 1/2x^-3/2So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2Once again simplifying gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xNeed help with Maths?One to one online tuition can be a great way to brush up on your Maths a Free Meeting with one of our hand picked tutors from the UK’s top universitiesFind a tutor W tym nagraniu wideo omawiam metodę rozwiązywania równań trygonometrycznych i pokazuję jak najlepiej rysować wykresy sinusa i nagrania: 25 \(2\sin x+3\cos x=6\) w przedziale \((0,2\pi )\) ma rozwiązań rzeczywistych. dokładnie jedno rozwiązanie rzeczywiste. dokładnie dwa rozwiązania rzeczywiste. więcej niż dwa rozwiązania rzeczywiste. ARozwiąż równanie \(\sin6x + \cos3x = 2\sin3x + 1\) w przedziale \(\langle 0, \pi \rangle\).\(x = 0, x = \frac{2}{3}\pi , x = \frac{7}{18}\pi, x = \frac{11}{18}\pi.\)Rozwiąż równanie \(\cos 3x+\sin 7x=0\) w przedziale \(\langle0,\pi\rangle\).\(x\in \left\{\frac{3}{8}\pi,\frac{7}{8}\pi,\frac{3}{20}\pi,\frac{7}{20}\pi,\frac{11}{20}\pi,\frac{15}{20}\pi,\frac{19}{20}\pi\right\}\)Rozwiąż równanie \((\cos x) \Biggl[ \sin \biggl(x - \frac{\pi}{3} \biggl) + \sin \biggl(x + \frac{\pi}{3} \biggl)\Biggl] = \frac{1}{2}\sin x\). \(x \in \biggl\{-\frac{\pi}{3} + 2k\pi, k\pi, \frac{\pi}{3} + 2k\pi\biggl\}\)Rozwiąż równanie \( \sqrt{3}\cdot \cos x=1+\sin x \) w przedziale \( \langle 0, 2\pi \rangle \) . \(x=\frac{3\pi }{2}\) lub \(x=\frac{\pi }{6}\)Dane jest równanie \(\sin x = a^2 + 1\), z niewiadomą \(x\). Wyznacz wszystkie wartości parametru \(a\), dla których dane równanie nie ma rozwiązań.\(a\in \mathbb{R} \backslash \{0\}\)Wyznacz, w zależności od całkowitych wartości parametru \(a\gt 0\), liczbę różnych rozwiązań równania \(\sin (\pi ax)=1\) w przedziale \(\left\langle 0,\frac{1}{a} \right\rangle \).Rozwiąż równanie \(\sin 2x+2\sin x+\cos x+1=0\), dla \(x\in \langle -\pi ,\pi \rangle \).\(-\frac{5\pi }{6}\), \(-\frac{\pi }{6}\), \(-\pi \), \(\pi \)Wyznacz wszystkie wartości parametru \(\alpha \in \langle 0;2\pi \rangle \), dla których równanie \((x^2-\sin 2\alpha )(x-1)=0\) ma trzy rozwiązania.\(\alpha \in (0;\frac{\pi }{4})\cup (\frac{\pi }{4},\frac{\pi }{2})\cup (\pi ;\frac{5\pi }{4})\cup (\frac{5\pi }{4};\frac{3\pi }{2})\)Wyznacz wszystkie wartości parametru \(a\), dla których równanie \((\cos x+a)\cdot (\sin^{2} x-a)=0\) ma w przedziale \(\langle 0,2\pi \rangle \) dokładnie trzy różne rozwiązania.\(a=1\)Rozwiąż równanie \(\sin \left(x+\frac{\pi}{6}\right)+\cos x=\frac{3}{2}\) w przedziale \(\langle 0; 2\pi \rangle \). \(x\in \left\{0, \frac{\pi}{3}, 2\pi \right\}\)Dana jest funkcja \(f(x)=\cos x\) oraz funkcja \(g(x)=f\left(\frac{1}{2}x\right)\). Rozwiąż graficznie i algebraicznie równanie \(f(x)=g(x)\). \(x=\frac{4}{3}k\pi \land k\in \mathbb{Z} \)Rozwiąż równanie \(\sin x|\cos x|=0,25\), gdzie \(x\in \langle 0; 2\pi \rangle\).\(x=\frac{\pi }{12}\) lub \(x=\frac{5\pi }{12}\) lub \(x=\frac{7\pi }{12}\) lub \(x=\frac{11\pi }{12}\)Rozwiąż równanie \(\cos2x + 2 = 3\cos x\).\(x=\frac{\pi }{3}+2k\pi \) lub \(x=-\frac{\pi }{3}+2k\pi \) lub \(x=2k\pi \) gdzie \(k\in \mathbb{Z} \)Rozwiąż równanie \(\cos 2x + \cos x + 1 = 0\) dla \(x\in \langle 0,2\pi \rangle\).\(x=\frac{\pi }{2}\) lub \(x=\frac{3\pi }{2}\) lub \(x=\frac{2\pi }{3}\) lub \(x=\frac{4\pi }{3}\)Rozwiąż równanie \(\cos 2x+3\cos x=-2\) w przedziale \(\langle 0,2\pi \rangle \). \bold{\mathrm{Basic}} \bold{\alpha\beta\gamma} \bold{\mathrm{AB\Gamma}} \bold{\sin\cos} \bold{\ge\div\rightarrow} \bold{\overline{x}\space\mathbb{C}\forall} \bold{\sum\space\int\space\product} \bold{\begin{pmatrix}\square&\square\\\square&\square\end{pmatrix}} \bold{H_{2}O} \square^{2} x^{\square} \sqrt{\square} \nthroot[\msquare]{\square} \frac{\msquare}{\msquare} \log_{\msquare} \pi \theta \infty \int \frac{d}{dx} \ge \le \cdot \div x^{\circ} (\square) |\square| (f\:\circ\:g) f(x) \ln e^{\square} \left(\square\right)^{'} \frac{\partial}{\partial x} \int_{\msquare}^{\msquare} \lim \sum \sin \cos \tan \cot \csc \sec \alpha \beta \gamma \delta \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega A B \Gamma \Delta E Z H \Theta K \Lambda M N \Xi \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \sin \cos \tan \cot \sec \csc \sinh \cosh \tanh \coth \sech \arcsin \arccos \arctan \arccot \arcsec \arccsc \arcsinh \arccosh \arctanh \arccoth \arcsech \begin{cases}\square\\\square\end{cases} \begin{cases}\square\\\square\\\square\end{cases} = \ne \div \cdot \times \le \ge (\square) [\square] ▭\:\longdivision{▭} \times \twostack{▭}{▭} + \twostack{▭}{▭} - \twostack{▭}{▭} \square! x^{\circ} \rightarrow \lfloor\square\rfloor \lceil\square\rceil \overline{\square} \vec{\square} \in \forall \notin \exist \mathbb{R} \mathbb{C} \mathbb{N} \mathbb{Z} \emptyset \vee \wedge \neg \oplus \cap \cup \square^{c} \subset \subsete \superset \supersete \int \int\int \int\int\int \int_{\square}^{\square} \int_{\square}^{\square}\int_{\square}^{\square} \int_{\square}^{\square}\int_{\square}^{\square}\int_{\square}^{\square} \sum \prod \lim \lim _{x\to \infty } \lim _{x\to 0+} \lim _{x\to 0-} \frac{d}{dx} \frac{d^2}{dx^2} \left(\square\right)^{'} \left(\square\right)^{''} \frac{\partial}{\partial x} (2\times2) (2\times3) (3\times3) (3\times2) (4\times2) (4\times3) (4\times4) (3\times4) (2\times4) (5\times5) (1\times2) (1\times3) (1\times4) (1\times5) (1\times6) (2\times1) (3\times1) (4\times1) (5\times1) (6\times1) (7\times1) \mathrm{Radians} \mathrm{Degrees} \square! ( ) % \mathrm{clear} \arcsin \sin \sqrt{\square} 7 8 9 \div \arccos \cos \ln 4 5 6 \times \arctan \tan \log 1 2 3 - \pi e x^{\square} 0 . \bold{=} + Related » Graph » Number Line » Similar » Examples » Our online expert tutors can answer this problem Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Your first 5 questions are on us! You are being redirected to Course Hero I want to submit the same problem to Course Hero Correct Answer :) Let's Try Again :( Try to further simplify Number Line Graph Hide Plot » Sorry, your browser does not support this application Examples \sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi \cos (x)-\sin (x)=0 \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi 2\sin ^2(x)+3=7\sin (x),\:x\in[0,\:2\pi ] 3\tan ^3(A)-\tan (A)=0,\:A\in \:[0,\:360] 2\cos ^2(x)-\sqrt{3}\cos (x)=0,\:0^{\circ \:}\lt x\lt 360^{\circ \:} trigonometric-equation-calculator cos^{2}x+2cosx+1=0 en Profile Edit Profile Messages Favorites My Updates Logout User qa_get_logged_in_handle sort Home Class 10th What is the domain of the function cos^-1 (2x –... User qa_get_logged_in_handle sort What is the domain of the function cos^-1 (2x – 3) Home Class 10th What is the domain of the function cos^-1 (2x –... by Chief of LearnyVerse (321k points) asked in Class 10th Mar 23 30 views What is the domain of the function cos^-1 (2x – 3)(a) [-1, 1](b) (1, 2)(c) (-1, 1)(d) [1, 2] 1 Answer Related questions

cos 2x 1 2