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Math Help - Sine and Cosine question

  1. #1
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    Sine and Cosine question

    I am learning sine, cosine and trigonometry graphs, and other trig stuff. And I am really stuck with this question:

    1) What is the value of x???
    2sinx + cosx = 1.5 (x is between 0 and 180)


    I would really appreciate it if someone could help me with this question.
    Thank You ^^
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Mathsaholic View Post
    I am learning sine, cosine and trigonometry graphs, and other trig stuff. And I am really stuck with this question:

    1) What is the value of x???
    2sinx + cosx = 1.5 (x is between 0 and 180)


    I would really appreciate it if someone could help me with this question.
    Thank You ^^
    If you have to solve an equation of the form a\sin x+b\cos x=c, you can divide the whole equation by \sqrt{a^2+b^2}. The new coefficients of \sin x and \cos x have the property that the sum of their squares is 1. This allows you to apply the Addition Formula for the \sin, which says that \sin(x)=\sin(x)\cos(x_0)+\cos(x)\sin(x_0).
    Then you use the two equations \cos(x_0)=a/\sqrt{a^2+b^2} and \sin(x_0)=b/\sqrt{a^2+b^2}, to figure out what x_0 is.

    So in your case this works as follows
    2\sin(x)+\cos(x)=1.5
    \frac{2}{\sqrt{5}}\sin(x)+\frac{1}{\sqrt{5}}\cos(x  )=\frac{1.5}{\sqrt{5}}
    \sin(x+x_0)=\frac{1.5}{\sqrt{5}}
    Now you apply \sin^{-1} to both sides to get the general solution
    x+x_0\approx \begin{cases}42.13^\circ+n\cdot360^\circ\\137.87^\  circ+n\cdot 360^\circ\end{cases}, n\in \mathbb{Z}

    Finally you have that \cos(x_0)=\frac{2}{\sqrt{5}} and \sin(x_0)=\frac{1}{\sqrt{5}}, in other words \tan(x_0)=\sin(x_0)/\cos(x_0)=\frac{1}{2}, which means, x_0\approx 26.57^\circ (we don't need the general solution for x_0 here).

    Thus you have

    x\approx \begin{cases}42.13^\circ-26.57^\circ+n\cdot360^\circ\\137.87^\circ-26.57^\circ+n\cdot 360^\circ\end{cases}, n\in \mathbb{Z}
    Now, from these infinitely many solutions you need to pick those, that lie in the reqired intervall.
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