# Math Grade 11 identities and equationsSolve 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.` **

sorry the actual question is: 2 - 2sin^2(θ) = cos(θ)