solve over the requested interval.

how do you solve these?

sin2x+2tan2x=0 (interval = negative infinity to positive infinity)

2sin^2(3x)-cos3x-1=0 same interval as the first one

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- May 3rd 2009, 06:06 PMsmyleeface17precal 5.1 to 5.3
solve over the requested interval.

how do you solve these?

sin2x+2tan2x=0 (interval = negative infinity to positive infinity)

2sin^2(3x)-cos3x-1=0 same interval as the first one - May 4th 2009, 04:25 AMstapel
Convert the tangent to sines and cosines. Multiply through by cos(2x) to clear the denominator. Factor, and solve the resulting trig equations.

Convert the squared sine to 1 - cos^2(3x). Then factor the resulting quadratic, and solve the factors.

If you get stuck, please reply showing how far you have gotten. Thank you! (Wink)