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How do I solve: 3cosx+4sinx+3=0

How do I solve: 3cosx+4sinx+3=0 Where x is between 0 & 360

I'm not sure how to begin !

Thanks,
Ross
Reply 1
Avatar for cj_134
cj_134
9
try expressing it in the form of Rcos(x+/-a) or Rsin(x+/-a)
Reply 2
Avatar for SalBhatti
SalBhatti
13
I think iteration is the only way to solve this kind of question. I'm in AS so I don't know that how iteration is applied, but you may search on google for it.
ive just finished this chapter in C2 last week and i think it wud be helpful if u used the formula: sinx/cosx = tanx

so it would be 4/3tanx + 3=0
then make tanx the subject and find tan-1 in ur calculator and then u can use the quadrant rule or the graph to help u find the angles
Original post by cj_134
try expressing it in the form of Rcos(x+/-a) or Rsin(x+/-a)


Zishi gave you the correct answer for people with C3 knowledge.

Another approach for those with a little more knowledge is to use a half tangent substitution.
Reply 5
Avatar for ztibor
ztibor
10
Original post by Rossblunsden
How do I solve: 3cosx+4sinx+3=0 Where x is between 0 & 360

I'm not sure how to begin !

Thanks,
Ross


THere is more way to solve this equation.
For example you can solve it graphically.

It seems most simple for me to use trigonometric identity of
sin2x+cos2x=1sin^2x+cos^2x=1
and from this change sinx with
sinx=1−cos2xsinx=\sqrt{1-cos^2x}
And arrange the equation
41−cos2x=3(1−cosx)4\sqrt{1-cos^2x}=3(1-cosx)
Take square both side and solve for cosx


Another method is the use of another tigonometric identity.
That is the addition rule
sin(A+x)=sinAcosx+cosAsinxsin(A+x)=sinAcosx+cosAsinx
We don't know sinA and cosA but we know from the equation that
the ratio of them is 3/4 that is
sinA=3tsinA=3t
cosA=4tcosA=4t
Taking square both equation and adding them
sin2A+cos2A=25t2sin^2A+cos^2A=25t^2
1=25t21=25t^2
t=15t=\frac{1}{5}
So sinA=3/5 cosA=4/5 and tanA=3/4 -->A
Divide the original equation by 5 and substitute
\frac{3}{5}cosx+\frac{4}{5}sinx+\frac{3}{5}=0
sinAcosx+cosAsinx=−35sinAcosx+cosAsinx=-\frac{3}{5}
sin(A+x)=−35sin(A+x)=-\frac{3}{5}


THe third method maybe that arranging the equation you use the identity of
sin2x2=1−cosx2sin^2\frac{x}{2}=\frac{1-cosx}{2} and
sinx=2sinx2cosx2sinx=2sin\frac{x}{2}cos\frac{x}{2}
THe equation
4sinx=3(1−cosx)4sinx=3(1-cosx)
8sinx2cosx2=6sin2x28sin\frac{x}{2}cos\frac{x}{2}=6sin^2\frac{x}{2}
Arranging to 0 and taking out sinx/2 as factor it gives the solutions


The 4th method would be writing up sine and cosine with t=tanx/2
t=tanx2t=tan\frac{x}{2}
sin2x2+cos2x2=1sin^2\frac{x}{2}+cos^2\frac{x}{2}=1
dividing by cosine squared
tan2x2+1=1cos2x2tan^2\frac{x}{2}+1=\frac{1}{cos^2\frac{x}{2}}
11+t2=cos2x2\frac{1}{1+t^2}=cos^2\frac{x}{2}
sin2x2=1−cos2x2=t21+t2sin^2\frac{x}{2}=1-cos^2\frac{x}{2}=\frac{t^2}{1+t^2}
So
cosx=cos2x2−sin2x2=1−t21+t2cosx=cos^2\frac{x}{2}-sin^2\frac{x}{2}=\frac{1-t^2}{1+t^2}
sin2x=4sin2x2⋅cos2x2=4t2(1+t2)2sin^2x=4sin^2\frac{x}{2}\cdot cos^2\frac{x}{2}=\frac{4t^2}{(1+t^2)^2}
that is
sinx=2t1+t2sinx=\frac{2t}{1+t^2}
Substituting them to the original equation we get a quadratic for t.


The fifth method maybe if use the identity of
±1+tan2x=1cosx\pm \sqrt{1+tan^2x}=\frac{1}{cosx}
Arranging the equation and dividing by cosx (cosx=0 would be examined as solution)
3cosx+4sinx=−33cosx+4sinx=-3
3+4tanx=−3cosx=−31+tan2x3+4tanx=\frac{-3}{cosx}=-3\sqrt{1+tan^2x}
So we've got a quadratic equation for tanx


Another method would be using complex valued, so
exponential functions.
cosx=cosh(ix)=12(eix+e−ix)cosx=cosh(ix)=\frac{1}{2}(e^{ix}+e^{-ix})
sinx=1isinh(ix)=12i(eix−e−ix)sinx=\frac{1}{i}sinh(ix)=\frac{1}{2i}(e^{ix}-e^{-ix})
This give quadratic for e^(ix).
(edited 13 years ago)

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