2016年4月1日 · x=pi/2+kpi, k in ZZ In the trigonometric circle you will notice that cos(x)=0 corresponds to x=pi/2 and also x=-pi/2. Additionally to these all the angles that make a …

2016年5月15日 · 90^o x= cos^-1(0) = 90^o Using the cosine graph, x could also = 270^o, 450^o, 810^o, -90^o, -270^o, -450^o, -810^o etc.

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

2016年2月7日 · How do you solve #sin(x/2) + cosx =0# in the interval [0, 2pi]? Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations 1 Answer

2016年2月24日 · Solution set is {pi/2, (7pi)/6, (3pi)/2, (11pi)/6} In 2sinxcosx+cosx=0, taking cosx common we get cosx(2sinx+1)=0 Hence, either cosx=0 whose solution in domain [0 ...

2018年7月9日 · Here, #2cosxsinx-cosx=0# #=>cosx(2sinx-1)=0# #=>cosx=0 or 2sinx-1=0# #=>cosx=0 or sinx=1/2# #(i)cosx=0=>color(blue)(x=(2k+1)pi/2 ,kin ZZ#

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

2016年12月16日 · General Solution for 2cos^2x+cosx=0 is x=((2n+1)pi)/2 or x=2npi+-(2pi)/3, where n is an integer. 2cos^2x+cosx=0 hArrcosx(2cosx+1)=0 i.e. either cosx=0 or 2cosx+1=0 …

2016年11月25日 · x = pi/2, (3pi)/2, pi/6, (5pi)/6 Before we solve, we need to note an identity: sin2x = 2sinxcosx Using the identity from above, rewrite the equation. 2sinxcosx - cosx=0 Now …

2018年4月5日 · sin 2x + cos x = 0 2sin x.cos x + cos x = 0 cos x(2sin x + 1) = 0 either factor should be zero. a. cos x = 0 Unit circle gives 2 solutions -->