Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 months ago

Can anyone solve it please?

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3 Answers

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  • 9 months ago
    Best Answer

    Let

    y=log(sin^2(x)+3)[cos(2x)+3], then

    (sin^2(x)+3)^y=cos(2x)+3

    =>

    yln[sin^2(x)+3]=ln[cos(2x)+3]

    =>

    y=ln[cos(2x)+3]/ln[sin^2(x)+3]

    =>

    y=ln[4-2sin^2(x)]/ln[3+sin^2(x)]

    Hint: sketch the graph of y to find

    where y>1 along the x-axis.

    Alternatively,

    y>1=>

    ln[4-2sin^2(x)]>ln[3+sin^2(x)]

    =>

    4-2sin^2(x)>3+sin^2(x)

    =>

    3sin^2(x)-1<0

    =>

    [sqr(3)sin(x)-1][sqr(3)sin(x)+1]<0

    =>

    -1/sqr(3)<sin(x)<1/sqr(3)

    =>

    -0.61548<x<0.61548

    =>|x|<0.61548

    or

    pi+0.61548>pi-x>pi-0.61548

    =>3.75707>pi-x>2.52611

    {the principal values}

    Check: take x=0.609

    LHS=ln[4-2(0.32723837...)]=1.20762311..

    RHS=ln[3+0.32723837...]=1.20214264..

    =>LHS>RHS, valid.

    take x=2.54

    LHS=ln[4-2(0.32030645..)]=1.21175854..

    RHS=ln[3+0.32030645..]=1.20005708..

    =>LHS>RHS, valid.

    take x=3.7

    LHS=ln[4-2(0.28072633..)]=1.23504909..

    RHS=ln[3+0.28072633..]=1.18806484..

    =>LHS>RHS, valid.

  • 9 months ago

    This is equivalent to [ln(cos 2x + 3)]/[ln(sin² x + 3)] > 1

    Then ln(cos 2x + 3) > ln(sin² x + 3), { note: sin² x ≥ 0 and ln(3) > 0 so no sign change necessary }

    and cos 2x + 3 > sin² x + 3

    cos 2x > sin² x

    1 - 2sin² x > sin² x

    1 > 3 sin² x

    sin² x < 1/3

    | sin x | < (√3)/3

    -0.61548 < x + kπ < 0.61548 (rounded) where k = an integer

  • rotchm
    Lv 7
    9 months ago

    Hint: log_a (b) = ln(b)/ln(a).

    Apply that. Take care in the signs and the domains.

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