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Math Help - Solve (3 sin x + 4 cos x)^2 = 3^2

  1. #1
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    Solve (3 sin x + 4 cos x)^2 = 3^2

    I need to solve this equation (3 sin x + 4 cos x)^2 = 3^2.

    I have solved this far, and it doesn't seem that I'm getting anywhere to the solution.

    9 sin^2 x + 24 sin x cos x + 16 cos^2 x = 9
    9 sin^2 x + 12 sin 2x + 16 cos^2 x = 9
    9 sin^2 x + 12 sin 2x + 16(1 - sin^2 x) = 9
    9 sin^2 x + 12 sin 2x + 16 - 16 sin^2 x = 9
    - 7 sin^2 x + 12 sin 2x = -7

    Where do I go from here?! It doesn't look I can factorise...
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  2. #2
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    Solve (3 sin x + 4 cos x)^2 = 3^2-img.jpg
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  3. #3
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    Thanks so much! Finally got this solved. So, x has two values = 90 deg, 343.7 deg (or -16.3 deg)!
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  4. #4
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    You're welcome!
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  5. #5
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    Quote Originally Posted by darkvelvet View Post
    Thanks so much! Finally got this solved. So, x has two values = 90 deg, 343.7 deg (or -16.3 deg)!
    + 360^{\circ}n where n is an integer.
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  6. #6
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    Hello, darkvelvet!

    \text{Solve: }\;(3\sin x + 4\cos x)^2 \:=\: 3^2

    Take square roots: . 3\sin x + 4\cos x \:=\:\pm3

    Divide by 5: . \frac{3}{5}\sin x + \frac{4}{5}\cos x \;=\;\pm\frac{3}{5} .[1]


    Let \,\theta be an angle in a right triangle with: . opp = 4,\;adj = 3,\;hyp = 5
    . . Then: . \sin\theta = \frac{4}{5},\;\cos\theta = \frac{3}{5},\;\tan\theta = \frac{4}{3}


    Substitute into [1]: . \cos\theta\sin x  + \sin\theta\cos x \:=\:\pm\frac{3}{5}

    Then we have: . \sin(x + \theta) \;=\;\pm\frac{3}{5}

    . . . . . . . . . . . . . . . x + \theta \;=\;\arcsin\left(\pm\frac{3}{5}\right)

    . . . . . . . . . . . . . . . . . x \;=\;\arcsin\left(\pm\frac{3}{5}\right) - \theta

    . . . . . . . . . . . . . . . . . x \;=\;\arcsin\left(\pm\frac{3}{5}\right) - \arcsin\left(\frac{4}{5}\right)
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