# solve equation on the interval 0<theta<2pisin(theta)-sqroot(3)cos(theta)=1

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Solve `sintheta-sqrt(3)costheta=1` on `(0,2pi)` :

`sin^2theta-2sqrt(3)sinthetacostheta+3cos^2theta=1` after squaring both sides

`sin^2theta+cos^2theta-2sqrt(3)sinthetacostheta+2cos^2theta=1` but `sin^2theta+cos^2theta=1` so

`-2sqrt(3)sinthetacostheta+2cos^2theta=0`

`2costheta(-sqrt(3)sintheta+costheta)=0`

Since we squared we must check for extraneous solutions:

`2costheta=0 ==> theta = pi/2,(3pi)/2` but `(3pi)/2` yields -1 in the original equation.

`-sqrt(3)sintheta+costheta=0`

`==>costheta=sqrt(3)sintheta`

`==>tantheta=1/sqrt(3)`

`==>theta=pi/6,(7pi)/6` but substituting `pi/6` into the original equation yields -1=1.

**So the two solutions are `theta=pi/2,(7pi)/6` **

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The graph:

You need to use the fundamental formula of trigonometry, such that:

`sin^2 theta + cos^2 theta = 1 => sin theta = sqrt(1 - cos^2 theta)`

Substituting `sqrt(1 - cos^2 theta)` for sin theta yields:

`sqrt(1 - cos^2 theta) - sqrt(cos theta) = 1`

`sqrt(1 - cos^2 theta) = 1 + sqrt(cos theta)`

Raising to square both sides yields:

`1 - cos^2 theta = 1 + 2sqrtcos theta + cos theta`

Reducing duplicate terms yields:

`-cos^2 theta = 2sqrtcos theta + cos theta`

You need to move the terms to one side such that:

`2sqrtcos theta + cos theta + cos^2 theta = 0 `

Factoring out `sqrt cos theta` yields:

`sqrt cos theta(2 + sqrt cos theta + cos theta*sqrt cos theta) = 0`

`sqrt cos theta = 0 => cos theta = 0 => theta = {pi/2; (3pi)/2}`

`2 + sqrt cos theta + cos theta*sqrt cos theta = 0`

You need to come up with the substitution such that: `sqrt cos theta = t => cos theta = t^2`

`2 + t + t^3 = 0 => t^3 + t + 1 + 1 = 0`

`(t^3 + 1) + (t + 1) = 0 => (t + 1)(t^2 - t + 1) + (t + 1) = 0`

Factoring out `t + 1` yields:

Since `(t^2 - t + 1 + 1) > 0 => t + 1 = 0 => t = -1`

Substituting back `sqrt cos theta` for t yields:

`sqrt cos theta = -1` invalid

**Hence, evaluating the solutions to the given equation yields **`theta = {pi/2; (3pi)/2}.`