Solve for all values of x in the range 0° to 360°
7sin2x = cosx
Ive got as far as using a common trig identity of sin2A = 2sincosA
7(2sinθ cosθ) = cosθ
14sinθcosθ = cosθ
14sinθcosθ - cosθ = 0
cosθ(14sin-1)=0
either cosθ = 0
or 14sinθ -1 = 0
sinθ = 1/14 = 0.071
cosˆ-1(0) = 90
θ= 90,270
sinˆ-1(0.071) = 4.071
θ= 4.071 and 360-4.071 = 355.929
Could someone please confirm i have done this the correct way if not please advise.
Thanks in advance.
Solve for all values of x in the range 0° to 360° the equation
7sin2x=cosx
Could you show me how to do this please.
I have started by finding the indetity of sin2A = 2sincosA but dont know what to do after that.
Also Help with this:
Solve the equation 7cos2θ = 3sinθ for all values of θ in the range 0° to 360°.
Once again i have selected the trig identity but where to go from there i am struggling.
Thanks
Hello,
So for the first one :
7sin(2x)=cos(x)
14sin(x)cos(x)-cos(x)=0
cos(x)[14sin(x)-1]=0
and you know that a product ab=0 if and only if either a=0 or b=0 (or both)
for the second one, use the trig identity for cos(2x) involving only sin(x) :
cos(2x)=1-2[sin(x)]^2
so the equation is now :
7(1-2sin²(x))=3sin(x)
7-14sin²(x)-3sin(x)=0
let t=sin(x) and solve the quadratic equation
No need to use so many tags you know
As you said sin(2x) = 2sin(x)cos(x) so 7sin(2x) = 14sin(x)cos(x) = cos(x)
You could divide by cos(x) and solve but that won't give you all the solutions, instead take cos(x) from both sides and factorise
cos(x)(14sin(x)-1) = 0
then either cos(x) = 0 in which case x = 90, 270 or
14sin(x)-1 = 0, sin(x) = 1/14 in which case x = 4.10 (3sf)
so your solutions are 4.10, 90 and 270
---------
Please do not double post
http://www.mathhelpforum.com/math-he...x-0-360-a.html