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mukushla

  • 3 years ago

Solve this equation\[\sin^6 x+\cos^6 x=\frac{5}{8}\]for \(x \in[0,2\pi]\)

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  1. sauravshakya
    • 3 years ago
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    sin^6 x + cos^6 x (sin^2x)^3+(cos^2x)^3 (sin^2x + cos^2x) (sin^4x-sin^2x cos^2x +cos^4x) 1*{(sin^2x +cos^2x)^2 -3sin^2x cos^2x} 1-3sin^2xcos^2x 1-3(sinx cosx)^2 1-3(sin2x/2)^2 1-3/4 sin^2(2x)

  2. sauravshakya
    • 3 years ago
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    1-3/4 sin^2 (2x) =5/8 3/4 sin^2 (2x)=3/8 sin^2 (2x)=1/2 2sin^2 (2x) =1 1-2sin^2 (2x)=0 cos4x=0

  3. sauravshakya
    • 3 years ago
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    Now LET a=4x then, cos(a)=0 , 0<=a<=8pi

  4. sauravshakya
    • 3 years ago
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    Solve for a

  5. sauravshakya
    • 3 years ago
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    I hope this will lead to the solution.

  6. sauravshakya
    • 3 years ago
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    And remember x=a/4

  7. sauravshakya
    • 3 years ago
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    Is that correct @mukushla

  8. mukushla
    • 3 years ago
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    nice job :)

  9. harsh314
    • 3 years ago
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    i think if you do it like this it would be shorter \[ (1-\cos^{2}x)^{ 3}\]\[(1-\cos^{2}x)^{ 3} +\cos ^{6}x=\frac{ 5}{ 8 }\] \[1-\cos ^{6}x-3\cos ^{2}x(1-\cos ^{2}x)+\cos ^{6}x=\frac{ 5 }{ 8 }\] the two cos^6 terms cancel \[-3\cos ^{2}x \times \sin ^{2}x=\frac{ 5 }{ 8 }-1\] \[3\sin ^{2}x \times \cos ^{2}x=\frac{ 3 }{ 8 }\] multiply both sides by 4/3 and we have\[4\sin ^{2}xcos ^{2}x=\frac{ 1 }{ 2 }\] and 2sinx cosx=sin2x so \[\sin ^{2}2x=\frac{ 1 }{ 2 }\] hence \[2x=\frac{ \pi }{ 4 } or 2x=\frac{ 3\pi }{ 4 }\] hence x is pi/8 or 3pi/8

  10. harsh314
    • 3 years ago
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    it is looking cumbersome but if you work it out it wont be so.........

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