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anushkaudeshan

  • 4 years ago

3sec^4 x - 10sec^2 x - 8 = 0, what is sin x

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  1. estudier
    • 4 years ago
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    Set sec^2 as y, solve quadratic for sec2, sub for 1 + tan^2 etc...

  2. MathsBoy
    • 4 years ago
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    \[(3\sec^{2}(x) - 4)(\sec^{2}(x) - 2) = 0\] \[\sec^{2}(x) = \frac{4}{3} \space\vee\space \sec^{2}(x) = 2\] \[\cos^2(x) = \frac{3}{4} \space\vee\space \cos^2(x) = \frac{1}{2}\] \[1-\sin^{2}(x) = \frac{3}{4} \space\vee\space 1-\sin^{2}(x) = \frac{1}{2}\] \[\sin(x) = \sqrt{\frac{1}{4}} \space\vee\space \sin(x) = \sqrt{\frac{1}{2}}\]

  3. MathsBoy
    • 4 years ago
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    I'm very sorry, I made a very stupid mistake: \[(3\sec^2(x) + 2)(\sec^2(x) - 4) = 0\] \[\sec^2(x) = -\frac{2}{3} \space\vee\space \sec^2(x) = 4\] \[\cos^2(x) = -\frac{3}{2} \space\vee\space \cos^2(x) = \frac{1}{4}\] \[\sin^2(x) = \frac{5}{2} \space\vee\space \sin^2(x) = \frac{3}{4}\] There is no real inverse sine of 5/2, so only carry on with the latter: \[\sin(x) = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}\] Please accept my apologies for the incorrect answer.

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