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solve trig equation involving sin2x=2sinxcosx in a given interval

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First replace $sin2x$ with $2sinxcosx$ and rearrange

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\\begin{align}

\\var{a}sin2x & =\\simplify{0+{b} cosx}  \\\\\\\\
\\var{a}\\times 2sinxcosx & =\\simplify{0+{b} cosx} \\\\\\\\
\\simplify{2{a}}sinxcosx \\simplify{0-{b}cosx} & =0 \\\\\\\\
cosx(\\simplify{2{a}}sinx-\\var{b}) &=0
\\end{align}

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This can now be solved to give

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\\[ cosx=0 \\text{  or  } sinx=\\frac{\\var{b}}{\\simplify{2{a}}} \\]

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Taking $cosx=0$ and remembering that for $cosx$ a second angle can be found by talking the first away from $2 \\pi$

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\\begin{align}

cosx=0 \\implies x=cos^{-1}(0)=\\frac{\\pi}{2} \\text{  or  } \\frac{3 \\pi}{2}

\\end{align}

\n

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Next taking $cosx=\\frac{\\var{b}}{\\simplify{2{a}}}$ and remembering that for $sinx$ a second angle can be found by talking the first away from $ \\pi$

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\\begin{align}

sinx=\\frac{\\var{b}}{\\simplify{2{a}}} \\implies x=sin^{-1}\\left(\\frac{\\var{b}}{\\simplify{2{a}}}\\right)=\\var{x1} \\text{  or  } \\var{x3}

\\end{align}

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To 3 significant figures and in radians the answers are therefore

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$\\var{x1_3} \\text{  and  } \\var{x2_3} \\text{  and  } \\var{x3_3} \\text{  and  } \\var{x4_3}$

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Solve the equation $\\var{a}sin2x=\\simplify{0+{b} cosx}$ in the range $0\\leq x \\leq 2\\pi$

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

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$x_1=$[[0]]

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$x_2=$[[1]]

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$x_3=$[[2]]

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$x_4=$[[3]]

\n

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The first setp is to replace $sin2x$ with $2sinxcosx$

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