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5 votes
Solve over the interval [0, 2pi]

sin^2x = cosx + 1

User Ghaul
by
8.0k points

1 Answer

3 votes

Answer:

The solutions are

S

=

{

1

2

π

,

3

2

π

,

1

6

π

,

5

6

π

}

Explanation:

We need

sin

2

x

=

2

sin

x

cos

x

Therefore,

sin

2

x

=

cos

x

sin

2

x

cos

x

=

0

2

sin

x

cos

x

cos

x

=

0

cos

x

(

2

sin

x

1

)

=

0

So,

{

cos

x

=

0

2

sin

x

1

=

0

,

{

cos

x

=

0

sin

x

=

1

2

,

{

x

=

π

2

3

2

π

x

=

1

6

π

5

6

π

x

[

0

,

2

π

]

The solutions are

S

=

{

1

2

π

,

3

2

π

,

1

6

π

,

5

6

π

}

graph{sin(2x)-cosx [-1.622, 9.475, -2.51, 3.04]}

User Solitud
by
7.1k points