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Byju's Answer
Standard XII
Mathematics
Principal Solution of Trigonometric Equation
Number of sol...
Question
Number of solution(s) of the equation
ln
(
2
−
sin
2
x
)
=
0
where
x
∈
[
0
,
10
π
]
is
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Solution
ln
(
2
−
sin
2
x
)
=
0
⇒
2
−
sin
2
x
=
1
⇒
sin
2
x
=
1
=
sin
2
π
2
⇒
x
=
n
π
±
π
2
Now,
x
∈
[
0
,
10
π
]
⇒
x
=
π
2
,
3
π
2
,
5
π
2
,
⋯
,
19
π
2
Number of solutions
=
10
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0
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Principal Solution of Trigonometric Equation
Standard XII Mathematics
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