Answer:
Step-by-step explanation:
Answer:
x
=
π
3
,
2
π
3
,
4
π
3
,
5
π
3
Explanation:
(
sin
x
)
2
=
3
(
cos
x
)
2
(
sin
x
)
2
=
3
(
1
−
(
sin
x
)
2
)
(
sin
x
)
2
=
3
−
3
(
sin
x
)
2
4
(
sin
x
)
2
=
3
(
sin
x
)
2
=
3
4
sin
x
=
±
(
√
3
2
)
x
=
π
3
,
π
−
π
3
,
π
+
π
3
,
(
2
π
)
−
π
3
x
=
π
3
,
2
π
3
,
4
π
3
,
5
π
3
If this was in the region
0
≤
x
≤
2
π
[tex]\bf \textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin^2(x)=3cos^2(x)\implies sin^2(x)=3[1-sin^2(x)] \implies sin^2(x)=3-3sin^2(x) \\\\\\ sin^2(x)+3sin^2(x)=3\implies 4sin^2(x)=3\implies sin^2(x)=\cfrac{3}{4}[/tex]
[tex]\bf sin(x)=\pm\sqrt{\cfrac{3}{4}}\implies sin(x)=\pm\cfrac{\sqrt{3}}{\sqrt{4}}\implies sin(x)=\pm\cfrac{\sqrt{3}}{2} \\\\\\ sin^{-1}[sin(x)]=sin^{-1}\left( \pm\cfrac{\sqrt{3}}{2} \right)\implies x= \begin{cases} \frac{\pi }{3}\\\\ \frac{2\pi }{3}\\\\ \frac{4\pi }{3}\\\\ \frac{5\pi }{3} \end{cases}[/tex]
that is, on the interval [0, 2π].
