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 ...

2017年2月18日 · x = pi/2, (3pi)/2, pi/4, (3pi)/4, (5pi)/4 and (7pi)/4 Note that cos3x can be rewritten as cos(2x + x). cosx + cos(2x + x) = 0 Now use cos(A + B) = cosAcosB - sinAsinB. cosx + cos2xcosx - sin2xsinx = 0 Apply cos2x = 2cos^2x -1 and sin2x = 2sinxcosx. cosx + (2cos^2x - 1)cosx - 2sinxcosx(sinx) = 0 cosx + 2cos^3x - cosx - 2sin^2xcosx = 0 Use sin^2x + cos^2x = 1: …

2018年4月27日 · (3pi)/4; (7pi)/4 Use trig identity. sin x + cos x = sqrt2cos (x - pi/4) = 0 cos (x - pi/4) = 0 Unit circle gives 2 solutions: a. x - pi/4 = pi/2 x = pi/2 + pi/4 ...

2015年8月12日 · How do you solve #\cos^2 x = \frac{1}{16} # over the interval #[0,2pi]#? How do you solve for x in #3sin2x=cos2x# for the interval #0 ≤ x < 2π# See all questions in Solving Trigonometric Equations

2016年9月4日 · x = pi/2 and (3pi)/2 Apply the identity cos(3x) = cos(2x + x): cos(2x + x ) = 0 We can now expand using the sum formula cos(A + B)= cosAcosB - sinAsinB. cos2xcosx + sin2xsinx = 0 Use the following double angle identities to expand further: •cos2alpha = 1 - 2sin^2alpha •sin2alpha = 2sinalphacosalpha ~~~~~ (1 - 2sin^2x)cosx + 2sinxcosxsinx = 0 cosx - …

2015年6月4日 · f(x) = cos x(2cos x + 1) = 0. Solve the 2 basic trig equations: cos x = 0 and #cos x = -1/2#. Within period #(0, 2pi):#

2016年9月26日 · x=0, pi and 2pi. This is a cubic equation in cos x. Sum of the coefficients is 0. So, cos x = 1 is a solution

2016年7月13日 · The domain is (-infty, infty) and the range is [-1,1]. This is a fun problem because we are presented with a modified version of the cos(x) function, but as we will see, it is, in fact, not in any way different from the standard version. cos|x| is the cos(x) function with the absolute value of x input into it. What this means is that if x >= 0 then it is replaced with x. If x …

2018年4月16日 · sinx=0, cosx=-1, tanx=0, cotx=undefined secx= -1 If cosecant is undefined, that means it is over 0. Remember that cosecant is hypotenuse over opposite side (reciprocal of sine). Hypotenuse is 1, so sine should be 0/1, or 0. sin^2x+cos^2x=1 0^2x+cos^2x=1 cos^2x=1 cosx=+-1 rarr Disregard the positive 1 because the question says that cosx<0. cosx=-1 tanx=sinx/cosx …

2018年6月18日 · 2npi+-cos^(-1)(0.6) normally you can get x=cos^(-1)(0.6) using cos(-x)=cos(x) x is also -cos^(-1)(0.6) Because cos(x+2npi)=cos(x)(n=integer number) general solutions ...