Question

2 sin x cos X - COS X
+ x
1-sin x + sin- x - COS
2
X
= cotx​

Answer 1

Answer:

okhhhhhhhhhh its hard

Answer 2

Step-by-step explanation:

Given, 2(sinx−cos2x)−sin2x(1+2sinx)+2cosx=0

2{sinx−(1−2sin

2

x)}−{2sinxcosx(1+2sinx)}+2cosx=0

............. [∵sin2x=2sinxcosx]and(∵1−cos2x=2sin

2

x)

2{sinx−(1−2sin

2

x)}+2cosx{1−sinx(1+2sinx)}=0

or (2sin

2

x+sinx−1)(1−cosx)=0

or (sinx+1)(2sinx−1)(1−cosx)=0

∴sinx=−1,cosx=1,sinx=

2

1

∴x=2nπ+

2

3π

,x=2nπ,nπ+(−1)

n

6

π