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Math Help - Trig inequality

  1. #1
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    Trig inequality

    Although we haven't done them in school yet, I would like someone to show me how to solve it!!

    \cos^2 \frac{x}{3} \le \sin^2 \frac{x}{3} - \frac{1}{2}
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  2. #2
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    Hello, livmed!

    \cos^2\!\left(\frac{x}{3}\right) \;\le \;\sin^2\!\left(\frac{x}{3}\right) - \frac{1}{2}

    We have: . \cos^2\!\left(\frac{x}{3}\right)- \sin^2\left(\frac{x}{3}\right) \;\leq \;-\frac{1}{2}

    . . Hence: . . . . . . . \cos\left(\frac{2x}{3}\right) \;\leq\;-\frac{1}{2}


    We find that: . \cos\theta\:\leq\:-\frac{1}{2} when \theta is between \frac{2\pi}{3} and \frac{4\pi}{3}


    So we have: . \frac{2\pi}{3} \;\leq \;\frac{2x}{3} \;\leq\;\frac{4\pi}{3}


    Multiply by \tfrac{3}{2}\!:\;\;\boxed{\pi \;\leq \;x \;\leq \;2\pi }

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