A Calculus problem. Please show steps.?

Question

See this,
http://www.geocities.com/black_rx20022002/math_problem.JPG
Solve by using double-angle, half-angle formulas, whatever.

*Don't know how to input the square thingy, so i made it as a picture.. pretty silly...

Anyway, please do not skip any steps please.
Appreciated.

Answer 1

The answers so far are incomplete.

Let y = sin(2x)
2y^2 + y - 1 = 0
(2y - 1)(y + 1) = 0
so y = 1/2 or y = -1
and since y = sin(2x),
sin(2x) = 1/2 or sin(2x) = -1

If sin(2x) = 1/2
2x = π/6, 5π/6, 13π/6, 17π/6
(you have to go around the circle twice since we will be dividing these values in half
[i.e, if 0 ≤ 2x ≤ 4π , then 0 ≤ x ≤ 2π])
and x = π/12, 5π/12, 13π/12, 17π/12

if sin(2x) = -1
2x = 3π/2 or 7π/2
(again here, we went around the circle twice to find values for 2x, since we will be dividing these in half)
and x = 3π/4, 7π/4

so x = π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4

And the picture is not silly at all, I use them all the time.
However, if you wanted a superscript 2, you could use the character map in windows, or hold down the "Alt" key and type 253 on the number pad, then release the Alt key. Also, sin(x) squared is often denoted sin^2(x).

Answer 2

2sin²2x + sin2x - 1 = 0
sin2x = (-1 ± √(1+8))/4
sin2x = (-1 ± √(9))/4
sin2x = (-1 ± 3)/4
sin2x = -1, +1/2
2x = -90°, 30°
x = -45°, 15°

Answer 3

Not really a calc problem.

2sin²2x + sin2x - 1 = 0

u = sin2x

2u² + u - 1 = 0
(2u - 1)(u + 1) = 0

u = 1/2 or u = -1

sin2x = 1/2
2x = arcsin 1/2
x = 1/2 arcsin 1/2

sin2x = -1
2x = arcsin -1
x = 1/2 arcsin -1

Answer 4

remember, sin(x)/x as x is going to 0 is a million. tan(x)^2 / x => sin(x)^2 / (x * cos(x)^2) => (sin(x)/x) * sin(x) / cos(x)^2 x is going to 0 a million * sin(0) / cos(0)^2 => a million * 0 / a million => 0 Or, shall we use L'hopital's rule, which says that for indeterminate types The decrease of f(x)/g(x) is comparable to the decrease of f'(x)/g'(x) f(x) = tan(x)^2 f'(x) = 2 * tan(x) * sec(x)^2 g(x) = x g'(x) = a million tan(0)^2 / 0 => 0 / 0 this is indeterminate, so we are in a position to apply L'hopital's rule: 2 * tan(x) * sec(x)^2 / a million => 2 * tan(x) * sec(x)^2 x is going to 0 2 * tan(0) * sec(0)^2 => 2 * 0 * a million => 0

Answer 5

2sin²2x + sin2x - 1 = 0

So (2sin2x - 1)(sin2x + 1) = 0

sin2x = ½, -1 (0 ≤ x ≤ 2π → 0 ≤ 2x ≤ 4π)

So 2x = π/6, 5π/6, 13π/6, 17π/6, 3π/2, 7π/2

So x = π/12, 5π/12, 13π/12, 17π/12, 3π/4, 7π/4
= π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4

Answer 6

2sin^2(2x)+sin(2x)-1=0
u=sin(2x)
2u^2+u-1=0
(2u-1)(u+1)=0
u=1/2 or u=-1
sin(2x)=1/2 or sin(2x)=-1
we know that sin(pi/6)=sin(5pi/6)=1/2
so x=pi/12 or 5pi/12
we know sin(3pi/2)=-1
so x=3pi/4
in total, x=pi/12,5pi/12, and 3pi/4