Please find all solutions. sin2x - cos2x = -1?

Question

Find all solutions..please help

Answer 1

2sinxcosx - 1 + 2sin^2 x = - 1 .



2sinx(cosx - sinx) = 0 .



sinx = 0 or cosx - sinx = 0 .



When sinx = 0, x = npi .



When cosx - sinx = 0 , cosx = sinx ,



tanx = 1 ,



x = npi + pi/4 .



Since, trigonometric functions are dependent on angles and such angles can take any value, we cannot specify a particular value for such a function, unless it is specified that the angle assumes values within a particular closed or an open interval.



For example rotation of minute and hour hands of a clock from their initial positions. It can take values even higher than 2pi.



Thus, general solution for x is (npi) and (npi + pi/4).



A few possible values for this particular variable are x = 0, pi, pi/4, 5pi/4, etc.



Remember, we have to solve for x and the value of angle x is important.

Answer 2

both 2sin xcos x+cos^2 x-sin^2 x=a million 2sin xcos x+cos^2 x-sin^2 x=sin^2 x+cos^2 x 2sin^2 x=2sinxcosx 2sin^2 x-2sinxcosx=0 2sinx(sinx-cosx)=0 sin x=0 OR sin x=cos x x=0,one hundred and eighty x=40 5,one hundred thirty 5 or 2sin xcos x+cos^2 x-sin^2 x=-a million 2sin xcos x+cos^2 x-sin^2 x=-sin^2 x-cos^2 x -2cos^x=2 sin x cos x 2 sin x cos x+2cos^x=0 2cosx(sin x+cox x)=0 cos x=0 OR sin x+cos x=0 x=ninety x=-40 5,-one hundred thirty 5 therefore x=0,one hundred and eighty,40 5,one hundred thirty 5,ninety,-40 5,-one hundred thirty 5 i imagine it is the reply.Please tell no matter if that is faulty.

Answer 3

Using some trig identities..



sin(2x) = 2sin(x)cos(x)



And -1+cos2x = -1(1-cos(2x)) = -2sin^2(x)



So, 2sin(x)cos(x) = -2sin^2(x) -> cos(x) = -sin(x) -> -1 = tan(x)...



So, when is -1 = tan(x), well..that's where sin = cos, but one is negative. And that's at 3pi/4 + n*pi, where n is an integer.



To check....



sin(2(3pi/4) - cos(2(3pi/4)) = -1 -> sin(3pi/2) - cos(3pi/2) = -1 -0 = -1



sin(2(7pi/4) - cos(2(7pi/4)) = -1 -> sin(7pi/2) - cos(7pi/2) = -1 - 0 = -1



etc...



Hope that helps.