Question

if x belongs to [-2π,2π] and log0.5sinx=1-log0.5cosx then find the solution of the equation ​

Answer 1

Answer:

log

0.5

sinx=1−log

0.5

cosx where x∈[−2π,2π]

Above equation is valid when, sinx>0,

⇒x∈(−2π,−π)∪(0,π)

and cosx>0

⇒x∈(−2π,−

2

3π

)∪(−

2

π

,

2

π

)∪(

2

3π

,2π)

Therefore, x∈(−2π,−

2

3π

)∪(0,

2

π

)

log

0.5

sinx=1−log

0.5

cosx

⇒log

0.5

sinx+log

0.5

cosx=1

⇒log

0.5

(sinxcosx)=1,[∵loga+logb=log(ab)]

⇒sinxcosx=0.5

1

=

2

1

⇒2sinxcosx=1

⇒sin2x=1

⇒cos2x=0

⇒1−2sin

2

x=0

⇒sin

2

x=

2

1

⇒sinx=±

2

1

Now, ∵x∈(−2π,−

2

3π

)∪(0,

2

π

)

∴x={−

4

7π

,

4

π

}

Hence there are two solutions in the given interval

Ans: B