Solve for x; give solutions within the interval [0, 2π) 3 sinx = 1 + cos2x
B BlueStreak New member Joined May 29, 2019 Messages 16 May 30, 2019 #1 Solve for x; give solutions within the interval [0, 2π) 3 sinx = 1 + cos2x
Otis Elite Member Joined Apr 22, 2015 Messages 4,414 May 30, 2019 #2 Hello. What have you tried or thought about, so far? Hint: We can get an equation in quadratic form, by using a double-angle identity for cos(2x). ?
Hello. What have you tried or thought about, so far? Hint: We can get an equation in quadratic form, by using a double-angle identity for cos(2x). ?
B BlueStreak New member Joined May 29, 2019 Messages 16 May 30, 2019 #3 @Otis Would I convert cos(2x) to cos2x-sin2x? Last edited: May 30, 2019
Harry_the_cat Elite Member Joined Mar 16, 2016 Messages 3,693 May 30, 2019 #4 That wouldn't help much. Use the identity that involves sin only.
B BlueStreak New member Joined May 29, 2019 Messages 16 May 30, 2019 #5 @Harry_the_cat @Otis Thank you so much! I figured it out. My answers were π/6 and 5π/6. Appreciate the help! ?
@Harry_the_cat @Otis Thank you so much! I figured it out. My answers were π/6 and 5π/6. Appreciate the help! ?