College Trig ?

Find all solutions of the equation in the interval [0, 2pi]

(2cosx- square root of 2)(2sinx- square root of 3)

write your answer in radians in terms of pi.

if there is more than one solution, separate them with commas.

2 Answers

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  • TomV
    Lv 7
    3 weeks ago

    (2cos(x) - √2)(2sin(x) - √3)

    is an expression, not an equation.

    Equations have solutions, expressions do not.

    Rewrite the expression as an equation and it may be possible to find a solution.

    In order to be an equation, the expression MUST contain an equal, "=", sign.

    ==========================

    If we assume the expression is equal to zero (not necessarily true), then:

    (2cos(x) - √2)(2sin(x) - √3) = 0

    From the Zero Product Property, we find:

    In the interval [0, 2π]

    2cos(x) - √2 = 0 → cos(x) = √2/2 → x = π/4 (Q1), 7π/4 (Q4)

    {Since cosΘ > 0 in Q1 and Q4, there is a solution in Q1 and Q4}

    2sin(x) - √3 = 0 → sin(x) = √3/2 → x = π/3 (Q1), 2π/3 (Q2)

    {Since sinΘ > 0 in Q1 and Q2, there is a solution in Q1 and Q2}

    ANS:

    The set of values that individually satisfy the assumed equation are:

    x = π/4, π/3, 2π/3, 7π/4

  • 3 weeks ago

    I think you actually mean:

    (2cosx - √2)(2sinx - √3) = 0

    so, either cosx = √2/2 or sinx = √3/2

    Then, x = ±π/4 + 2nπ, π/3 + 2nπ or 2π/3 + 2nπ

    so, x = π/4, π/3, 2π/3 and 7π/4

    :)>

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