Thread: precal 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

Originally Posted by smyleeface17

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

Convert the tangent to sines and cosines. Multiply through by cos(2x) to clear the denominator. Factor, and solve the resulting trig equations.

Originally Posted by smyleeface17

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

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!