2016年5月24日 · #sinx# is known as a periodic function that oscillates at regular intervals. It crosses the x-axis (i.e. it is #0#) at #x = 0, pi,# and #2pi# in the domain #[0,2pi]#, and continues to cross the x-axis at every integer multiple of #pi#. graph{sinx [-10, 10, -5, 5]} And if you click on the graph, you get: So, whenever #sinx = 0#, we have that:

2015年2月3日 · The sine function equals 0 for every multiple of pi~~3.14 graph{sinx [0, 7, -1.2, 1.2]} pi radians is equal to 180^0 The sine-function goes from 0 to 1 at 90^0=pi//2 And then back to 0 at 180^0=pi Down to -1 at 270^0=3pi//2 And up to 0 again at 360^0=2pi So it will be 0 at every multiple of pi Summary sin x=0 ->x=k*pi, with k any whole number

2015年10月14日 · Hence sin2x=2sinx*cosx we have that sinx+sin2x=0=>sinx+2sinxcosx=0=>sinx(1+2cosx)=0=> sinx=0 or cosx=-1/2 ...

2017年4月18日 · Solve # sinx - cosx = 0#? Trigonometry Trigonometric Identities and Equations Solving Trigonometric ...

2015年6月7日 · cosx + sinx = 0 cos x = -sinx 1 = -tanx -1 = tanx tanx is equal to -1 at (3pi)/4 and (7pi)/4. Trigonometry .

2015年4月14日 · To solve a trig equation, transform it to a few basic trig equations. Solving trig equations finally results in solving basic trig equations. Reminder: 4 basic trig equations are: sin x = a; cos x = a; tan x = a; cot x = a. f(x) = sin 2x + sin x = 2sin xcos x + sin x = 0. f(x) = sin x(2cos x + 1 ) = 0. Next, use the trig unit circle and trig table to solve the 2 basic trig equations: sin x = 0 ...

Solving concept. To solve a trig equation, transform it into one, or many, basic trig equations. Solving a trig equation, finally, results in solving various basic trig equations.

2015年4月14日 · x=sin^-1(-1/2), x=sin^-1(1) Solution cos2x+sinx=0 As cos2x=cos^2x-sin^2x So cos^2x-sin^2x+sinx=0 1-sin^2x-sin^2x+sinx=0 1-2sin^2x+sinx=0 -2sin^2x+sinx+1=0 2sin^2x ...

2016年3月3日 · #lim_{x rarr 0} x/{sin x} = lim_{x rarr 0} 1/{cos x} = 1/{cos 0} = 1/1 = 1#. NOTE The question was posted in " Determining Limits Algebraically " , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem.

2018年3月17日 · #sin(2x)+sinx=0# #2sinxcosx+sinx=0# #sinx(2cosx+1)=0# #sinx=0,qquadcosx=-1/2# Here's a unit circle to remind us of where the sine and cosine values are: