sin x - cos 2x + 1 = 0?

Question

Find all solutions of this equation...
How do I find the solution? Please explain. Thanks

Thank you so much, richie_rich_abc!

Thanks Pascal! You've been really helpful

Answer 1

sin x - cos (2x) + 1 = 0
sin x - (cos² x - sin² x) + 1 = 0
sin x - (1 - 2 sin² x) + 1 = 0
2 sin² x + sin x = 0
(2 sin x + 1) sin x = 0

So either sin x = 0 or 2 sin x + 1 = 0. In the latter case:

2 sin x + 1 = 0
2 sin x = -1
sin x = -1/2

So either sin x equals 0 (in which case x is coterminal to either 0 or π) or sin x = -1/2 (in which case x is coterminal to either 7π/6 or 11π/6). As before, we must test these solutions:

sin 0 - cos 0 + 1 = 0 → 0 - 1 + 1 = 0, so x=0 is valid
sin π - cos (2π) + 1 = 0 → 0 - 1 + 1 = 0, so x=π is valid
sin (7π/6) - cos (14π/6) + 1 = 0 → -1/2 - 1/2 + 1 = 0, so x=7π/6 is valid
sin (11π/6) - cos (22π/6) + 1 = 0 → -1/2 -1/2 + 1 = 0, so x=11π/6 is valid

Ergo, all of the solutions we just found are valid, and x is coterminal to any one of 0, π, 7π/6, of 11π/6

Edit: You're welcome, Jessica.

Answer 2

First of all, you have to turn either sin x to cos x or turn cos x to sinx, and then solve by factorizing. One easy way is to use the double angle formula to turn cos 2x to sin x. The formula is

cos 2x = 1 - 2 (sin x)^2

Sub this to the original question,

sin x - [ 1 - 2(sin x)^2 ] + 1 = 0
sin x - 1 + 2(sin x)^2 + 1 = 0
2(sin x)^2 + sin x = 0

Take the common factor, sin x, out from the equation,

(sin x)( 2 sin x + 1) = 0

sin x = 0
x = 0 deg, 180 deg, and 360 deg.

or

2 sin x + 1 = 0
2 sin x = -1
sin x = -1/2
x = -30 deg (330 deg)

Using CAST rule, we know another solution is 180+30 = 210 deg.

Therefore, all the solutions are 0, 180, 210, 330, 360.

Answer 3

Use an identity for cos2x

cos2x = 1- 2sin^2(x)

sinx - (1 - 2sin^2(x)) + 1= 0

2sin^2(x) + sinx = 0

sinx(2sin^2(x) +1) = 0

sinx = -1/2 meaning x = 210º, 330º
or sinx = 0 meaning x = 0º or 180º