1°
cos
(
x
+
π
6
)
cos
(
x
−
π
6
)
=
1
2
⇔
cos
2
x
+
cos
π
3
=
1
⇔
cos
2
x
=
1
2
⇔
x
≡
±
π
6
mod
π
{\displaystyle \cos \left(x+{\frac {\pi }{6}}\right)\cos \left(x-{\frac {\pi }{6}}\right)={\frac {1}{2}}\Leftrightarrow \cos 2x+\cos {\frac {\pi }{3}}=1\Leftrightarrow \cos 2x={\frac {1}{2}}\Leftrightarrow x\equiv \pm {\frac {\pi }{6}}\mod \pi }
.
2°
sin
(
2
x
+
π
6
)
sin
(
2
x
−
π
6
)
=
1
2
⇔
cos
π
3
−
cos
4
x
=
1
⇔
cos
4
x
=
−
1
2
⇔
x
≡
±
π
6
mod
π
2
{\displaystyle \sin \left(2x+{\frac {\pi }{6}}\right)\sin \left(2x-{\frac {\pi }{6}}\right)={\frac {1}{2}}\Leftrightarrow \cos {\frac {\pi }{3}}-\cos 4x=1\Leftrightarrow \cos 4x=-{\frac {1}{2}}\Leftrightarrow x\equiv \pm {\frac {\pi }{6}}\mod {\frac {\pi }{2}}}
.
3°
cos
(
2
x
+
π
3
)
+
cos
(
2
x
−
π
3
)
=
−
3
2
⇔
2
cos
π
3
cos
2
x
=
−
3
2
⇔
x
≡
±
5
π
12
mod
π
{\displaystyle \cos \left(2x+{\frac {\pi }{3}}\right)+\cos \left(2x-{\frac {\pi }{3}}\right)=-{\frac {\sqrt {3}}{2}}\Leftrightarrow {\cancel {2\cos {\frac {\pi }{3}}}}\cos 2x=-{\frac {\sqrt {3}}{2}}\Leftrightarrow x\equiv \pm {\frac {5\pi }{12}}\mod \pi }
.
4°
cos
(
2
x
−
π
3
)
−
cos
2
x
=
3
2
⇔
2
sin
π
6
sin
(
2
x
−
π
6
)
=
3
2
⇔
2
x
−
π
6
∈
{
π
3
,
2
π
3
}
+
2
π
Z
⇔
x
≡
π
4
ou
5
π
12
mod
π
{\displaystyle \cos \left(2x-{\frac {\pi }{3}}\right)-\cos 2x={\frac {\sqrt {3}}{2}}\Leftrightarrow {\cancel {2\sin {\frac {\pi }{6}}}}\sin \left(2x-{\frac {\pi }{6}}\right)={\frac {\sqrt {3}}{2}}\Leftrightarrow 2x-{\frac {\pi }{6}}\in \{{\tfrac {\pi }{3}},{\tfrac {2\pi }{3}}\}+2\pi \mathbb {Z} \Leftrightarrow x\equiv {\frac {\pi }{4}}{\text{ ou }}{\frac {5\pi }{12}}\mod \pi }
.
5°
cos
(
x
+
π
3
)
sin
(
x
+
π
6
)
=
−
3
+
1
4
⇔
sin
(
2
x
+
π
2
)
=
sin
π
6
−
1
2
−
3
2
⇔
2
x
+
π
2
∈
{
−
π
3
,
−
2
π
3
}
+
2
π
Z
⇔
x
≡
−
5
π
12
ou
−
7
π
12
mod
π
{\displaystyle \cos \left(x+{\frac {\pi }{3}}\right)\sin \left(x+{\frac {\pi }{6}}\right)=-{\frac {{\sqrt {3}}+1}{4}}\Leftrightarrow \sin \left(2x+{\frac {\pi }{2}}\right)={\cancel {\sin {\frac {\pi }{6}}-{\frac {1}{2}}}}-{\frac {\sqrt {3}}{2}}\Leftrightarrow 2x+{\frac {\pi }{2}}\in \{-{\tfrac {\pi }{3}},-{\tfrac {2\pi }{3}}\}+2\pi \mathbb {Z} \Leftrightarrow x\equiv -{\frac {5\pi }{12}}{\text{ ou }}-{\frac {7\pi }{12}}\mod \pi }
.
6°
sin
(
x
+
π
3
)
sin
(
x
−
π
3
)
=
−
1
2
⇔
cos
2
π
3
−
cos
2
x
=
−
1
⇔
cos
2
x
=
1
2
⇔
x
≡
±
π
6
mod
π
{\displaystyle \sin \left(x+{\frac {\pi }{3}}\right)\sin \left(x-{\frac {\pi }{3}}\right)=-{\frac {1}{2}}\Leftrightarrow \cos {\frac {2\pi }{3}}-\cos 2x=-1\Leftrightarrow \cos 2x={\frac {1}{2}}\Leftrightarrow x\equiv \pm {\frac {\pi }{6}}\mod \pi }
.