If 0 < x < Pi, find all x which satisfy: cos^2 (x) + ((rt3 - 1) / 2) * cos (x) - (rt3 / 4) = 0?
If 0 < x < Pi, find all x which satisfy:
cos^2 (x) + ((rt3 - 1) / 2) * cos (x) - (rt3 / 4) = 0
Any help would be great thanks :)
Update:Btw there is such x that satisfies this. Cos(x) is between -1 and 1.
Enter the following equation: ((cos(x))^2)+((((sqrt(3))-1)/2)cos(x))-((sqrt(3))/4)
into your calculator again, or try this site (online graphing). For this website, just copy paste it in, and click "Eval" to see the graph. http://www.coolmath.com/graphit/
And forgot to say this, but I will need to know how to get to the answer without using a graphing calculator to look at the x-intercepts.
Thanks in advance :)
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no such x satify it, since cos(x)>1 on that interval.
you ought to show f(x)-sinx=0 for all pi/2<x<pi. observe that tanx is defined right here and tanx=sinx/cosx Then f(x)-sinx=sin^3 x + cos^3 x tan x-sinx=sinx(sin^2 x+cos^2 x-a million) yet sin^2 x+cos^2 x=a million for any cost of x, so we've =sinx(a million-a million)=sinx(0)=0. sinx=2/3; observe that when you consider that sine is non-end function we've that x is a cost interior the form pi/2<x<pi (there are different stages, yet that's sufficient). Now sin^2 x=4/9. Then a million-cos^2 x=4/9; rearranging we get cos^2 x = 5/9 cosx=sqrt(5)/3 or -sqrt(5)/3 yet we've that pi/2<x<pi and the cosine function is damaging in this selection, so we ought to have cosx=-sqrt(5)/3 Now, f(2x)=sin(2x) (by utilising the 1st workout) = 2sinxcosx=2(2/3)(-sqrt(5)/3)=-4sqrt(5)/9
yeah, if you graph the equation, you see the line never crosses the x-axis, meaning x never reaches 0.