Question

Solve the equation sin^2 x=3 cos ^2 x

Answer 1

`\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}`

`\bf sin(x)=\pm\sqrt{\cfrac{3}{4}}\implies sin(x)=\pm\cfrac{√(3)}{√(4)}\implies sin(x)=\pm\cfrac{√(3)}{2} \\\\\\ sin^(-1)[sin(x)]=sin^(-1)\left( \pm\cfrac{√(3)}{2} \right)\implies x= \begin{cases} (\pi )/(3)\\\\ (2\pi )/(3)\\\\ (4\pi )/(3)\\\\ (5\pi )/(3) \end{cases}`

that is, on the interval [0, 2π].

Answer 2

Step-by-step explanation:

Answer:

x

=

π

3

,

2

π

3

,

4

π

3

,

5

π

3

Step-by-step 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