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PlEASE help me with this!! I need it as soon as possible.......
Find the answer from 0 to 2pi.
When I slove this, I got
cosx+sinx=(1/ square root of 2).
then I don't know what to do the rest. Please help me.

  • pre-cal/trig -

    if you get to cosx+sinx=1/square root of 2, well we are doing identities in my pre-cal class and cosx+sinx always equals 1, so it would be 1/square root of 2.

  • pre-cal/trig -

    Sorry, cosx+ sinx is not equal to 1 (from the identites). But, it was like this:
    (sinx)^2+ (cosx)^2= 1

  • pre-cal/trig -

    Can Anybody answer this question for me???? Please???
    PlEASE help me with this!! I need it as soon as possible.......
    Find the answer from 0 to 2pi

  • pre-cal/trig -

    square both side:
    (cosx+sinx)^x=(1/square root of 2)
    then solve: it will work.

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