Solve the following on the interval of [0,2π)
tan2x-tan2xtan^2x=2
Answers
Answered by
0
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Step-by-step explanation:
Call tan x = t. Use the trig identity:
tan
2
t
=
2
t
1
−
t
2
2
t
1
−
t
2
=
t
→
2
t
=
t
(
1
−
t
2
)
=
t
−
t
3
→
t
3
+
t
=
0
t
(
t
2
+
1
)
=
0
−
→
t
=
tan
x
=
0
-> x = pi and x = 2pi.
Check.
x
=
π
→
tan
x
=
0
and
tan
2
x
=
tan
(
2
π
)
=
0
Correct.
Answered by
1
Answer:
Move
tan
(
x
)
to the right:
tan
(
2
x
)
=
tan
(
x
)
Substitute
2
tan
(
x
)
1
−
tan
2
(
x
)
for
tan
(
2
x
)
:
2
tan
(
x
)
1
−
tan
2
(
x
)
=
tan
(
x
)
Multiply by both sides by
1
−
tan
2
(
x
)
tan
(
x
)
2
=
1
−
tan
2
(
x
)
1
=
−
tan
2
(
x
)
tan
(
x
)
=
i
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