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trig equation that uses cos2x=c^2-s^s

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Solve $\\simplify{2{a}{d}+2{b}{c}}sinx=\\simplify{{a}{c}cos2x-{a}{c}-2{b}{d}}$

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The first step is to replace $cos2x$ with $1-2sin^2x$ as below

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\\[ \\simplify{2{a}{d}+2{b}{c}}sinx=\\simplify{{a}{c}(1-2sin^2x)-{a}{c}-2{b}{d}} \\]

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If you expand the brackets and rearrange you will get the following.

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\\[ \\simplify{ {a}{c}sin^2x +{a}{d}sinx+{b}{c}sinx+{b}{d}} \\]

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which factorises to

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\\[ (\\simplify{{a}sinx+{b}})(\\var{c}sinx+\\var{d})=0 \\]

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You can now take each bracket separately

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\\[ sinx=\\frac{\\var{absb}}{\\var{a}} \\text{  or  }  sinx=\\frac{-\\var{d}}{\\var{c}} \\]

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Taking each separately

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\\[ x=sin^{-1}\\frac{\\var{absb}}{\\var{a}} = \\var{x1} \\text{  or  } \\pi-\\var{x1} \\]

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\\[ x=sin^{-1}\\frac{-\\var{d}}{\\var{c}} = \\var{x3} \\text{  or  } \\pi-\\var{x3} \\]

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To 3 significant figures and in ascending order this gives

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\\[ x=\\var{x1_3} \\text{  or  }\\var{x2_3} \\text{  or  }\\var{x3_3} \\text{  or  }\\var{x4_3} \\]

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Solve $\\simplify{2{a}{d}+2{b}{c}}sinx=\\simplify{{a}{c}cos2x-{a}{c}-2{b}{d}}$ between $0<x<2\\pi$

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Give your answers to 3 significant figures and in ascending order.

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[[0]]

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[[1]]

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[[2]]

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[[3]]

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The first step is to replace $cos2x$ with $1-2sin^2x$

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