Question

Solve the equation on the

interval [0, 27).
4(sin x)2 - 2 = 0
X
.
7
[?] []? []T
x =
4) 4 4 4
Write your answer in increasing order. Enter the
number that belongs in the green box.
Enter

Answer 1

see explanation

Explanation:

note that (sinx)² = sin²x

4sin²x - 2 = 0 ( add 2 to both sides )

4sin²x = 2 ( divide both sides by 4 )

sin²x =
`(2)/(4)` =
`(1)/(2)` ( take square root of both sides )

sinx = ±
`\sqrt{(1)/(2) }` = ±
`(1)/(√(2) )`

sinx =
`(1)/(√(2) )` ( sinx > 0 , then x is in 1st/2nd quadrants )

x =
`(\pi )/(4)` , π -
`(\pi )/(4)` =
`(3\pi )/(4)`

sinx = -
`(1)/(√(2) )` ( sinx < 0 , then x is in 3rd/4th quadrants )

x = π +
`(\pi )/(4)` =
`(5\pi )/(4)` , 2π -
`(\pi )/(4)` =
`(7\pi )/(4)`

solutions are

`(\pi )/(4)` ,
`(3\pi )/(4)` ,
`(5\pi )/(4)` ,
`(7\pi )/(4)` in the interval [ 0, 2π )