Solve the equation sin^2(x) - sinxcosx = 6cos^2(x)for 0<=x<=360
I factorised it but cannot find the answer.
(sinx+2cosx)(sinx-3cosx)=0
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Solve the equation sin^2(x) - sinxcosx = 6cos^2(x)for 0<=x<=360
I factorised it but cannot find the answer.
(sinx+2cosx)(sinx-3cosx)=0
sin x + 2 cos x = 0 or sin x - 3 cos x = 0
sin x = -2 cos x or sin x = 3 cos x
tan x = -2 or tan x = 3
Solve for x
thank you thank you , i should have known such an easy question.
answer to previous question. 296.6, 71.6, 117, 251.56
I also want to ask another question, also an trigonometry equation. These are not homework questions or anything, its just that im studying this topic by myself.
solve 5 tanx - 3 cosx = 0
My working.
(5sinx)/cosx - 3cosx = 0
(5sinx-3cos^2(x))/cosx = 0
5sinx - 3cos^2x = 0
5sinx = 3cos^2x
sinx/cos^2x = 3/5
tanx/cosx = 3/5
tanx = 3 cosx = 5
x=71.6 or 251.56
Which is wrong, so i need some help.
5 sin x = 3 cos2 x
5 sin x = 3(1 - sin2 x)
5 sin x = 3 - 3 sin2 x
3 sin2 x + 5 sin x - 3 = 0
sin x = [-5 +/- \sqrt{25 + 36}]/6
right thanks.