Math Grade 11 Identities and equations Solve the following quadratic trig equations algebraically (without graphing technology) for the domain `0leq theta leq 360^@` . Use exact solutions whenever...
Math Grade 11 Identities and equations
Solve the following quadratic trig equations algebraically (without graphing technology) for the domain `0leq theta leq 360^@` . Use exact solutions whenever possible.
`6cos^2 theta + cos theta - 1 = 0`
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Expert Answers
calendarEducator since 2011
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You should also use the following alternative method such that:
`6 cos^2 theta = 1 - cos theta`
You need to substitute `2sin^2(theta/2)` for `1 - sin theta` such that:
`6 cos^2 theta = 2sin^2(theta/2)`
You should write `cos^2 theta` such that:
`cos^2 theta = cos^2 2(theta/2)`
Substituting `1 - 2 sin^2(theta/2)` for `cos^2 2(theta/2)` yields:
`6cos^2 2(theta/2) = 2sin^2(theta/2) => 6(1 - 2 sin^2(theta/2)) = 2sin^2(theta/2)`
Reducing by 2 both sides yields:
`3(1 - 2 sin^2(theta/2)) = sin^2(theta/2)`
Opening the brackets yields:
`3 - 6sin^2(theta/2) = sin^2(theta/2)`
`3 = 7sin^2(theta/2) => sin^2(theta/2) = 3/7`
`sin (theta/2) = +-sqrt(3/7) ` `theta/2 = arcsin(+-sqrt(3/7)) => theta = 2sin(+-sqrt(3/7)) => theta = +-2sin(sqrt(3/7)) `
You may use graphical method to find the solutions to the give equation, such that:
Hence, evaluating the solutions to the given equation yields `theta = +-2sin(sqrt(3/7)).`
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calendarEducator since 2012
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First use substitution `cos theta=t.` Now equation becomes:
`6t^2-t-1=0` Now we solve quadratic equation
`t_(1,2)=(-(-1)pm sqrt(1-4cdot6cdot(-1)))/(2cdot6)=(1pm5)/(12)`
`t_1=-1/3`, `t_2=1/2`
So we have
`cos theta=-1/3=> theta=pm arccos(-1/3)+2k pi, k in ZZ` and
`cos theta=1/2=> theta={pi/3+2k pi,(2pi)/3+2k pi; k in ZZ}.`
So there are 4 solutions for `0leq theta leq 2pi` and they are:
`theta_1=arccos(1/3)`, `theta_2=-arccos(1/3)+2pi`, `theta_3=pi/3` and `theta_4=(2pi)/3`.
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