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Let’s check out your problem:
Petermine all solutions for
0
⩽
t
⩽
4
0 \leqslant t \leqslant 4
0
⩽
t
⩽
4
\newline
2
π
cos
(
π
t
)
+
2
π
cos
(
2
π
t
)
=
0
2 \pi \cos (\pi t)+2 \pi \cos (2 \pi t)=0
2
π
cos
(
π
t
)
+
2
π
cos
(
2
π
t
)
=
0
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Home
Math Problems
Grade 8
Solve multi-step inequalities
Full solution
Q.
Petermine all solutions for
0
⩽
t
⩽
4
0 \leqslant t \leqslant 4
0
⩽
t
⩽
4
\newline
2
π
cos
(
π
t
)
+
2
π
cos
(
2
π
t
)
=
0
2 \pi \cos (\pi t)+2 \pi \cos (2 \pi t)=0
2
π
cos
(
π
t
)
+
2
π
cos
(
2
π
t
)
=
0
Isolate Cosine Term:
Isolate one of the cosine terms.
\newline
2
π
cos
(
π
t
)
+
2
π
cos
(
2
π
t
)
=
0
2\pi \cos(\pi t) + 2\pi \cos(2\pi t) = 0
2
π
cos
(
π
t
)
+
2
π
cos
(
2
π
t
)
=
0
\newline
2
π
cos
(
π
t
)
=
−
2
π
cos
(
2
π
t
)
2\pi \cos(\pi t) = -2\pi \cos(2\pi t)
2
π
cos
(
π
t
)
=
−
2
π
cos
(
2
π
t
)
\newline
Divide both sides by
2
π
2\pi
2
π
.
\newline
cos
(
π
t
)
=
−
cos
(
2
π
t
)
\cos(\pi t) = -\cos(2\pi t)
cos
(
π
t
)
=
−
cos
(
2
π
t
)
Apply Double Angle Identity:
Use the double angle identity for cosine.
\newline
cos
(
π
t
)
=
−
cos
(
2
π
t
)
\cos(\pi t) = -\cos(2\pi t)
cos
(
π
t
)
=
−
cos
(
2
π
t
)
\newline
cos
(
π
t
)
=
−
1
+
2
sin
2
(
π
t
)
\cos(\pi t) = -1 + 2\sin^2(\pi t)
cos
(
π
t
)
=
−
1
+
2
sin
2
(
π
t
)
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\newline
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