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How do you solve sin2x = sinx?

source : socratic.org

How do you solve sin2x = sinx?

#sin 2x = sin x##2 sin x cos x = sin x##2 sin x cos x – sin x = 0##sin x (2 cos x – 1) = 0#Solution A: #sin x = 0 \Rightarrow x = kpi, k in ZZ#Solution B: #2 cos x = 1 \Rightarrow cos x = 1/2, x = pmpi/3+2kpi = pi/3(6kpm1), k in ZZ##therefore x=kpi# or #x = pi/3(6kpm1), k in ZZ#

Answered: sin(2x) - sin(x) = 0 | bartleby

Answered: sin(2x) – sin(x) = 0 | bartleby – Solution for sin(2x) – sin(x) = 0. Social Science. Anthropologysinx (2cosx – 1) =0. Now 2 things multiplied together are 0 one or both is 0. i.e. a*b =0 implies a or b is 0. In the above therefore either sinx =0 which implies x is 0 or 2cosx -1 =0Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step

How do you solve: sin2x – sinx = 0? | Yahoo Answers – Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.Transcript. Misc 7 Prove that: sin 3x + sin2x – sin x = 4 sin x cos 𝑥/2 cos 3𝑥/2 Solving L.H.S sin 3x + sin 2x − sin x = sin 3x + (sin 2x – sin x ) = sin 3x + 2cos ((2𝑥 + 𝑥)/2) . sin ((2𝑥−𝑥)/2) = sin 3x + 2cos (3𝑥/2) sin 𝑥/2 Using sin x – sin y = 2 cos (𝑥 + 𝑦)/2 sin (𝑥 − 𝑦)/2 Putting x = 2x & y = x , Rough As (3𝑥 + 𝑥)/2 = 4𝑥/2 = 2xSolve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.

How do you solve: sin2x - sinx = 0? | Yahoo Answers

2 sinx cosx= sin x – Step-by-Step Calculator – Symbolab – Squaring both sides gives 1+2\sin2x=\cos^2x-2\cos x\sin x+\sin^2x=1-\sin2x or \sin2x=0 This suggest x=k\pi/2 for k\in\mathbb{Z}, but care must be taken to eliminate the ones for which \cos x-\sin x=-1The problem statement What is the limit of sin2x/x as x approaches 0? Revelant equations lim sin(x)/x = 1 x–>0 Attempt at a solution so sin2x/x = 2sin(2x)/2x since sin(2x)/2x = 1 2sin(2x)/2x = 2*1 I know how to solve it this way however my teacher said you can solve it using double angle…sinx = sin2x. We know that sin2x = 2sinxcosx ==> sin x = 2sinx cosx ==> sinx – 2sinx*cosx = 0. Now factor sinx: ==> sinx (1- 2cosx) = 0 ==> sinx = 0 ==> x1 = 0, pi

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