Question

Solve sin(x)=cos(pi/5), where 3pi2 13 years ago

Answer 1

is that 3pi/2??

Answer 2

yes and its sin(x)=-cos(pi/5) forgot - infront of cos

Answer 3

okay..
cos pi/5 = -sin (3pi/2+ pi/5)
add 3pi/5 and pi/2, that is 17pi/10

Answer 4

okay

Answer 5

sinx=-sin(17pi/10)

Answer 6

sin(x) = -cos(pi/5)
sin(x) = -sin(pi/2 - pi/5)
sin(x) = sin(-(pi/2 - pi/5))
sin(x) = sin(pi/5 - pi/2)

use the identity :
sin(x) = sin(k*2pi + x)

so, the equation above can be
sin(x) = sin(k*2pi + pi/5 - pi/2)

cancel out the sin from both sides
x = sin(k*2pi + pi/5 - pi/2)
for k = 0, we get x = pi/5 - pi/2 (not satisfies)
for k = 1, we get x = 2pi + pi/5 - pi/2 =17pi/10
it is satisfies for x in the 3rd quadrant

Answer 7

i mean in the 4th quadrant :)

Answer 8

sin(x) = -sin(pi/2 - pi/5)
sin(x) = sin(-(pi/2 - pi/5))

instead of -sin(x)=sin(-(x)) can you use sin(pi+(x))?

Answer 9

i used the identity of sinx = cos(pi/2 - x)

Answer 10

no for the line -sin -> sin

Answer 11

oh, i used identity -sin(x) = sin(-x)

Answer 12

yes can u use -sin(x)=sin(pi+x) instead

Answer 13

yes, that will works too

Answer 14

how about when you used sin(x) = sin(k*2pi + x) , sin(x)=sin(pi-x+k^2pi) works too right

Answer 15

it's thought one

Answer 16

?? what thought one