Math Grade 11 identities and equations Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 ≤ x ≤ 360°. Use exact solutions whenever...
Math Grade 11 identities and equations
Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 ≤ x ≤ 360°. Use exact solutions whenever possible.
- 2 - 2sin2(θ) = cos(θ)
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You should use the fundamental formula of trigonometry `1 - cos^2 theta = sin^2 theta` such that:
`2 - 2(1 - cos^2 theta) = cos theta`
Opening the brackets yields:
`2 - 2 + 2cos^2 theta = cos theta`
Reducing like terms such that:
`2cos^2 theta = cos theta`
Moving all terms to one side yields:
`2cos^2 theta- cos theta = 0`
You should factor out `cos theta` such that:
`cos theta(2cos theta - 1) = 0`
You need to solve the following equations such that:
`cos theta = 0 => theta = pi/2 and theta = 3pi/2`
`2 cos theta - 1 = 0 => cos theta = 1/2 => theta = pi/6 and theta = 2pi - pi/6 => theta = 11pi/6`
Hence, evaluating the solutions to the given equation yields `theta = pi/6 , theta = pi/2 , theta = 3pi/2` and `theta= 11pi/6.`
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sorry the actual question is: 2 - 2sin^2(θ) = cos(θ)
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