cos(x) = − 1 2,1 cos ( x) = - 1 2, 1. Set up each of the solutions to solve for x x. cos(x) = − 1 2 cos ( x) = - 1 2. cos(x) = 1 cos ( x) = 1. Solve for x x in cos(x) = −1 2 cos ( x) = - 1 2. Tap for more steps x = 2π 3 +2πn, 4π 3 +2πn x = 2 π 3 + 2 π n, 4 π 3 + 2 π n, for any integer n n. 3 Answers. Sorted by: 8. One way will be to add it to the definition of Cos by Unprotect ing it. Unprotect [Cos] Cos [2 x] := 1 - 2 Sin [x]^2 Protect [Cos] Then evaluating the following: 2 Cos [2 x] + 3 x Cos [2 x] + Tan [x] Sin [x] Cos [2 x] + Exp [Tan [Cos [2 x]]] gives: Which you can further Simplify if you so please. High School Math Solutions – Trigonometry Calculator, Trig Identities. In a previous post, we talked about trig simplification. Trig identities are very similar to this concept. An identity Read More. Save to Notebook! Sign in. Free Pythagorean identities - list Pythagorean identities by request step-by-step. Vay Tiền Nhanh. You are using an out of date browser. It may not display this or other websites should upgrade or use an alternative browser. Forums Homework Help Precalculus Mathematics Homework Help What does (cos(2x))^2 equal? Thread starter justine411 Start date Apr 11, 2007 Apr 11, 2007 #1 Homework Statement (cos2x)^2Homework EquationsThe Attempt at a Solution I'm not sure if it is cos^2(2x) or cos^2(4x) or what. Should I use an identity to simplify it to make it easier to solve? Please help! :) Answers and Replies Apr 11, 2007 #2 What is there to solve??? (cos2x)^2 is just an expression. Apr 11, 2007 #3 In what sense is (cos(2x))2 a "problem"? What do you want to do with it? I will say that (cos(2x))2 means: First calculate 2x, then find cosine of that and finally square that result. Notice that it is still 2x, not 4x. The fact that 2 is outside the parentheses means that it only applies to the final result. Apr 12, 2007 #4 In what sense is (cos(2x))2 a "problem"? What do you want to do with it? I will say that (cos(2x))2 means: First calculate 2x, then find cosine of that and finally square that result. Notice that it is still 2x, not 4x. The fact that 2 is outside the parentheses means that it only applies to the final result. Doesn't (cos(2x))2 = cos2(2x)2 = cos2(4x2) ? Apr 12, 2007 #5 Doesn't (cos(2x))2 = cos2(2x)2 = cos2(4x2) ? No. 'Cos' is a particular operation and 2x is the argument. The exponent of 2 operates on cos, not on the argument. cos2y = cos y * cos y. There are also particular trigonometric identites with which one should be familiar, cos (x+y) and sin (x+y). Apr 12, 2007 #6 You still haven't told us what the problem was! Was it to write (cos(2x))^2 in terms of sin(x) and cos(x)? I would simply be inclined to write (cos(2x))^2 as cos^2(2x). Suggested for: What does (cos(2x))^2 equal? 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Register for no ads! \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 jak rozwiązać cosx = - 1/2 ja: jak rozwiązać równanie, gdy mamy, że cosx = −1/2 ogólnie, kiedy jest jakaś wartość ujemna gdyby było cosx = 1/2 to by było: x = π/3 + 2kπ lub x = − π/3 + 2kπ a jak robimy, gdy jest wartość ujemna? da się jakoś bez rysunku? ponoć pomocny jest wierszyk "w pierszej ćwiartce same plusy..." Bardzo proszę o pomoc 28 kwi 15:53 Jerzy: ⇔ cosx = −cos60 = cos(180 −60) = cos120 28 kwi 15:59 6latek : 28 kwi 16:00 ja: dziękuję bardzo 28 kwi 16:02 6latek : podobnie rozwiązujesz 28 kwi 16:03

cos 2x 1 2