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trig

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solve 2cos^2x-sinx=1, 0 < or equal to x < 2pie

  • trig -

    cos ^ 2 ( x ) = 1 - sin ^ 2 ( x )


    2 cos ^ 2 ( x ) - sin ( x ) = 1

    2 * [ 1 - sin ^ 2 ( x ) ] - sin ( x ) = 1

    2 - 2 sin ^ 2 ( x ) - sin ( x ) = 1 Subtract 1 to both sides

    2 - 2 sin ^ 2 ( x ) - sin ( x ) - 1 = 1 - 1

    - 2 sin ^ 2 ( x ) - sin ( x ) + 1 = 0


    Substitute :

    sin ( x ) = u


    - 2 u ^ 2 - u + 1 = 0


    The exact solution are :

    u = 1 / 2

    u = - 1


    cos ( x ) = 1 / 2

    cos ( x ) = - 1



    x = 30 ° = pi / 6

    x = 270 ° = 9 pi / 6 rad



    P.S

    If you don't know how to solve equation :

    - 2 u ^ 2 - u + 1 = 0

    In google type:

    quadratic equation online

    When you see list of results click on:

    Free Online Quadratic Equation Solver:Solve by Quadratic Formula

    When page be open in rectangle type:

    - 2 u ^ 2 - u + 1 = 0

    and click option: solve it!

    You wil see solution step-by-step

  • trig -

    Corection:

    sin ( x ) = 1 / 2

    sin ( x ) = - 1

  • trig -

    would the 3 answers be: pi/6, 5pi/6, 3pi/2

  • trig -

    Yes

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