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  1. MHB Apprentice

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    #1
    I have absolutely no idea how to tackle either of these questions

    Solve the following equation for angles between 0˚ and 360˚ to 2 decimal places
    4cosθ + 5sinθ = 3
    4 cot - 6 Cosec x = -6

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    #2
    Quote Originally Posted by simongreen93 View Post
    I have absolutely no idea how to tackle either of these questions

    Solve the following equation for angles between 0˚ and 360˚ to 2 decimal places
    4cosθ + 5sinθ = 3
    4 cot - 6 Cosec x = -6
    This has nothing to do with the hyperbolic functions!?

    To start you off:

    For the first, recall that $ \displaystyle cos^2( \theta ) = 1 - sin^2( \theta )$, so the equation becomes:
    $ \displaystyle 4 - 4~sin^2( \theta ) + 5~sin( \theta ) = 3$

    This is now a quadratic in $ \displaystyle sin( \theta )$

    For the second start with $ \displaystyle cot ( \theta ) = \frac{ cos( \theta )}{sin( \theta )}$ and $ \displaystyle cosec( \theta ) = \frac{1}{sin( \theta )}$

    -Dan

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    #3
    Quote Originally Posted by simongreen93 View Post
    I have absolutely no idea how to tackle either of these questions

    Solve the following equation for angles between 0˚ and 360˚ to 2 decimal places
    4cosθ + 5sinθ = 3
    4 cot - 6 Cosec x = -6
    First of all, these are trigonometric functions, not hyperbolic.

    Second, both are solved in the exact same way. Use the Pythagorean Identity (or some variation of it) to rewrite your equation all in terms of the same trigonometric function. Then solve the resulting quadratic equation.

    So in the first for example, you will substitute $\displaystyle \begin{align*} 1 - \sin^2{(\theta )} \end{align*}$ for $\displaystyle \begin{align*} \cos^2{(\theta )} \end{align*}$.

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