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Solving $\\sin(2x)-\\tan(x)=0$ for $x\\in \\left(0,\\frac{\\pi}{2}\\right)$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Solve $\\sin(2x)-\\tan(x)=0$, for $x \\in \\left(0,\\frac{\\pi}{2}\\right)$.
", "advice": "To solve \\[ \\sin(2x)-\\tan(x)=0, \\]
\nwe want to rewrite the the equation using the following trigonometric identities:
\n\\[ \\begin{split} \\sin(2x) - \\tan(x) &\\,=0 \\\\ 2\\sin(x)\\cos(x) - \\dfrac{\\sin(x)}{\\cos(x)} &\\,= 0 \\\\ 2\\sin(x)\\cos^2(x) - \\sin(x) &\\,= 0 \\\\ 2\\cos^2(x) - 1 &\\,=0. \\end{split} \\]
\nTherefore,
\n\\[ \\begin{split} \\cos^2(x) &\\,=\\frac{1}{2} \\\\ \\cos(x) &\\,= \\frac{1}{\\sqrt{2}}\\\\ x &\\, = \\cos^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right). \\end{split} \\]
\nSo our solution is \\[ \\begin{split} x &\\, = \\cos^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right) \\\\ &\\,= \\frac{\\pi}{4} \\\\ &\\,= 0.785 \\, (3\\text{ d.p.}). \\end{split} \\]
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