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    Hi
    Can someone kindly get me on to the right track, i am kinda lost maybe because i am doing it late lol. Anyways, the questions are: find general solution of 2cos^2x=1 .4sin^2 (2x)=3
    for the first one i thought of using the trig . identity but grr i keep getting wrong values.
    Thanks .
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    LOL another question, this ones bothering me real bad.
    solve: 1+sin(3x+1) = 0.3
    Well i started off by timing out the brackets:
    1+sin3x+sin= 0.3
    sin3x+sin = 0.3-1
    sin3x= sin^-1 (-0.7)
    sin3x = about -0.77539
    3x= sin^-1 (-0.77539)
    x= -0.8873/ 3 = which gives me -0.25 -.-
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    (Original post by ibysaiyan)
    LOL another question, this ones bothering me real bad.
    solve: 1+sin(3x+1) = 0.3
    Well i started off by timing out the brackets:
    1+sin3x+sin= 0.3
    sin3x+sin = 0.3-1
    sin3x= sin^-1 (-0.7)
    sin3x = about -0.77539
    3x= sin^-1 (-0.77539)
    x= -0.8873/ 3 = which gives me -0.25 -.-
    You can't 'multiply out' the bracket.
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    (Original post by gilbo)
    You can't 'multiply out' the bracket.
    Oh.. great now what, do i send it to the left or let x= -1/3?> hmm
 
 
 
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