solve equation on the interval 0<theta<2pisin(theta)-sqroot(3)cos(theta)=1
2 Answers
embizze | High School Teacher | (Level 2) Educator Emeritus
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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sciencesolve | Teacher | (Level 3) Educator Emeritus
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}.`