# Thread: solving for all values of x 0° to 360°

1. ## solving for all values of x 0° to 360°

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

2. Originally Posted by jamm89
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

I get [tex]sin^{-1}(1/14)= 4.096, not 4.071. Also, sin(360-x)= -sin(x), not sin(x). You want 180- x.

3. Originally Posted by HallsofIvy
I get [tex]sin^{-1}(1/14)= 4.096, not 4.071. Also, sin(360-x)= -sin(x), not sin(x). You want 180- x.
I think the 4.096 not 4.071 was due to rounding but ive changed it.

So am i right in saying the points that it crosses are 4.096, 90, 175.9 and 180?

4. ## Identity problem

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

5. Hello,
Originally Posted by jamm89
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

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

6. Originally Posted by jamm89
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.

[/FONT]
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

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7. For the first one i get
sinΘ= 1/14 = 0.071
and cosΘ=0 so cosˆ-1 (0)=90

then sinˆ-1(0.071) = 4.096

so Θ = 4.096, 90, (180-4.096) and 180

is that correct?