Question

Find all solutions in the interval [0, 2π).

4 sin2x - 4 sin x + 1 = 0

Answer 1

just consider sin x= y...you'll have 4 y^2-4 Y+1=0..solve for y first

Answer 2

y= -4-1

Answer 3

pi/6 is the first one right?

Answer 4

nope..solve either by factorizing or by quadratic equation and see what you get \[D= -b \pm \sqrt{b^2-4ac}/2a\]

Answer 5

yes it's pi/6

Answer 6

&& then 11pi/6

Answer 7

nope..then it's 2 pi /5

Answer 8

just use pi/6 + pi/2 \[\frac{ \pi }{ 6 }+ \frac{ \pi }{ 2 }\]

Answer 9

@babydoll332 did you get?

Answer 10

5pi/6

Answer 11

nope it's 4pi/6= 2pi/3

Answer 12

i can't find that 1 :/

Answer 13

take LCM of 2 and 6 you'll have 6 in the denominator and then \[\frac{ \pi }{ 6 }+\frac{ \pi }{ 2 }= \frac{ \pi+3*\pi }{ 6 }= \frac{ \pi +3\pi }{ 6 }= \frac{ 4\pi }{ 6 }= \frac{ 2 \pi }{ 3}\]

Answer 14

is ti clear?

Answer 15

see you have pi/6 and you need to find answer between (0,2pi) so just use 2 pi- pi/6