1+sinx/1-sinx=3. Find the value of x
Answers
Given,
(sinx−1)
3
+(cosx−1)
3
+(sinx)
3
=(2sinx+cosx−2)
3
----------------1
We know that, (a+b+c)
3
=a
3
+b
3
+c
3
+(a+b+c)(ab+bc+ca)
Here a=sinx−1,b=cosx−1,c=sinx
(a+b+c)=(2sinx+cosx−2)
(ab+bc+ca)=(2sin
2
x+2sinxcosx−3sinx−cosx+1)
(1)(sinx−1)
3
+(cosx−1)
3
+(sinx)
3
=(sinx−1)
3
+(cosx−1)
3
+(sinx)
3
+(2sinx+cosx−2)(2sin
2
x+2sinxcosx−3sinx−cosx+1)
⇒(2sinx+cosx−2)(2sin
2
x+2sinxcosx−2sinx−sinx−cosx+1)=0
⇒(2sinx+cosx−2)(2sinx−1)(sinx+cosx−1)=0
(i) sinx=
2
1
∴x=
6
π
,
6
5π
( as x∈[0,2π])
or
(ii) sinx+cosx=1
∴x=0,
2
π
,2π ( as x∈[0,2π])
or
(iii) 2sinx+cosx−2=0
2(sinx−1)=−cosx
4sin
2
x−8sinx+4=cos
2
x
5sin
2
x−8sinx+3=0
∴sinx=1or
5
3
cosx=2(1−sinx)
=2(1−
5
3
)=
5
4
sinx>0 and cosx>0
∴ x has one solution in 1st quadrant.
From (i),(ii),(iii), number of solutions of x=6
Answer:
take this
Step-by-step explanation:
= sin^2(x/2) +cos^2((x/2) + 2sin(x/2)*cos(x/2)
={sin(x/2) +cos(x/2)}^2
Similarly,
1-sinx
= sin^2(x/2)+cos^2(x/2) - 2sin(x/2)*cos(x/2)
={sin(x/2)- cos(x/2)}^2