trigonometric equations as maths help

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unserious
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#1
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#1
sqrt3 Sin(theta) = Cos(theta)
in the interval 0 < theta < 360

so far ive used the fact that sin over cos makes tan theta.

now ive been left with the multiplication of root 3 and tan(theta) = 0


I don't know where to go from here, all help appreciated
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I1Mu12
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#2
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As your dividing by cosz it will be tan(theta)=1 not 0
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hannanj0
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#3
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Is it root 3 multiplied by tan theta, or is it root 3 + tan theta?
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#4
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(Original post by hannanj0)
Is it root 3 multiplied by tan theta, or is it root 3 + tan theta?
multiplied
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#5
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(Original post by I1Mu12)
As your dividing by cosz it will be tan(theta)=1 not 0
where does the 1 come from?
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I1Mu12
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#6
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let x=theta (because I don't want to keep writing it)
sinx=cosx
sinx/cosx=cosx/cosx
tanx=1
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I1Mu12
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#7
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Then it would be 1/root 3 (I didn't see root 3 before)
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hannanj0
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#8
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(Original post by unserious)
multiplied
Oh in that case I would first get the root 3 on RHS, so you get sin theta = cos theta/root3
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mathstutor24
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#9
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\sqrt3 sin\theta=cos\theta

\frac{\sqrt3 sin\theta}{cos\theta }=\frac{cos\theta}{cos\theta}

\sqrt3 tan\theta=1

tan\theta =\frac{1}{\sqrt3}
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