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Math Help - Solving Trig Equation - Hidden restriction?

  1. #1
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    Solving Trig Equation - Hidden restriction?

    Hello everyone,

    I arrived at two solutions for the following trigonometric equation, but my textbook's answer only lists one of them.

    Could someone please check if I overlooked a restriction or erred during my solving process?

    Thank you very much!

    ---

    1. Solve  \sqrt{2} \sin x = \sqrt{3} - \cos x, 0 \leq x \leq 2\pi .

    ---

    I squared the equation to get:

     2 \sin^2 x = 3 - 2\sqrt{3}\cos x + \cos^2 x

    Applying Pythagorean Theorem:

     2(1 - \cos^2 x) = 3 - 2\sqrt{3}\cos x + \cos^2 x

     -3\cos^2 x + 2\sqrt{3} \cos x - 1 = 0

    Let  y = \cos x :

     -3y^2 + 2\sqrt{3}y - 1 = 0

    Since y = 0.5773;

    therefore:  \cos x = 0.5773

     x = 0.9554, 5.328

    However, the answer given is only (Scroll with mouse to highlight white text -->) 0.96.
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  2. #2
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    Quote Originally Posted by scherz0 View Post
    Hello everyone,

    I arrived at two solutions for the following trigonometric equation, but my textbook's answer only lists one of them.

    Could someone please check if I overlooked a restriction or erred during my solving process?

    Thank you very much!

    ---

    1. Solve  \sqrt{2} \sin x = \sqrt{3} - \cos x, 0 \leq x \leq 2\pi .

    ---

    I squared the equation to get:

     2 \sin^2 x = 3 - 2\sqrt{3}\cos x + \cos^2 x

    Applying Pythagorean Theorem:

     2(1 - \cos^2 x) = 3 - 2\sqrt{3}\cos x + \cos^2 x

     -3\cos^2 x + 2\sqrt{3} \cos x - 1 = 0

    Let  y = \cos x :

     -3y^2 + 2\sqrt{3}y - 1 = 0

    Since y = 0.5773;

    therefore:  \cos x = 0.5773

     x = 0.9554, 5.328

    However, the answer given is only (Scroll with mouse to highlight white text -->) 0.96.
    When you square an equation you introduce the possibility of extraneous solutions. All solutions obtained need to be checked by substituting in the original equation.
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