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haleyking345

  • 3 years ago

Solving trig Equations a.) sin^2x=3 cos^2x

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  1. haleyking345
    • 3 years ago
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    Please Help!!!

  2. jtvatsim
    • 3 years ago
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    reposting question in better format: \[\sin^2(x) = 3\cos^2(x)\]

  3. haleyking345
    • 3 years ago
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    I do not have a clue where to even start.

  4. jtvatsim
    • 3 years ago
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    well, as it stands it looks pretty bad... do you remember what sin(x)/cos(x) is equal to?

  5. haleyking345
    • 3 years ago
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    tan x

  6. jtvatsim
    • 3 years ago
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    good! now, let's use that to make this question simpler. divide both sides by cos^2(x) and it starts looking better.

  7. jtvatsim
    • 3 years ago
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    You should get this:\[\tan^2(x) = 3\]

  8. haleyking345
    • 3 years ago
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    So then you square root it right?

  9. jtvatsim
    • 3 years ago
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    yes!

  10. jtvatsim
    • 3 years ago
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    do not forget that you will have a positive and negative root

  11. haleyking345
    • 3 years ago
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    ok thank you

  12. jtvatsim
    • 3 years ago
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    Let me know if you need any further help good luck!

  13. SithsAndGiggles
    • 3 years ago
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    Alternatively, you can use the identity \[\sin^2x +\cos^2x=1\] Rewriting the left (or right side, appropriately), you have \[\sin^2x=3(1-\sin^2x)\\ \sin^2x=3-3\sin^2x\\ 4\sin^2x=3\\ \sin^2x=\frac{3}{4}\]

  14. SithsAndGiggles
    • 3 years ago
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    Which gives you the difference of squares \[\sin^2x-\left(\frac{\sqrt3}{2}\right)^2=0\\ \left(\sin x+\frac{\sqrt3}{2}\right)\left(\sin x-\frac{\sqrt3}{2}\right)=0\] Easily solvable.

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