Question

General solution of the equation |cosx|=sinx is

Answer 1

My ans is 2nπ( +- )π/4
Bt the ans is not in match with that given in ans book

Answer 2

thats correct,

-pi/4 is also +3pi/4

Answer 3

Bt the ans is [nπ+ (-1)^n π/4]

Answer 4

What i did was squared both lhs and rhs and what i ended up with was the expression
Cos2x=cosπ/2

Answer 5

\(\cos 2x = 1 \\ \cos 2x = (2n \pi + \pi/2 ) \\ x = ...\)
you'll get the book answer.

although i think both the answers are equivalent

Answer 6

\(\sin x \gt 0\) in I, II quadrants

Answer 7

\(\cos 2x = 1 \\ \cos 2x =\cos (2n \pi + \pi/2 ) \\ x = ...\)

Answer 8

Actually my first post was not right ...
It was li'l bit faulty in the sense that it should be nπ(+-)π/4

Answer 9

@hartnn
Bt cosx=cosy
Implies that x=2nπ(+-)y
Isn't it?

Answer 10

yes.

so thats why we get

2x = 2npi +- pi/2 here

Answer 11

Yep bt if we put the value as 5π/4 into the original equation we'll not get it to be matching with the thing given in front of it...
How??

Answer 12

because while squaring both sides,
we have introduced solutions that are not actually the part of original equation.

Answer 13

So what's the better way?

Answer 14

for example,
if |x| = y

x^2 =y^2
x = +y, x=-y

but the original equation is not true when x=1, y=-1

Answer 15

better way:

|x| = y
implies
x = y or
x =-y (y >0)

Answer 16

U mean that i should open the mod sign and solve it individually...

Answer 17

The two equations would come and i need to take their intersection ...?

Answer 18

I m not getting it still..
I opened the mod sign...and came up with this
x=2nπ+π/4
And x=2nπ+3π/4

Answer 19

x can't be 2nπ+3π/4
because sin x will become negative

Answer 20

5pi/4, 7pi/4 ..they all make sin x as negative

Answer 21

yes sin (2npi -pi/4) is negative

Answer 22

How to do this question in a proper sequence?

Answer 23

U there??