sin2x-cosx=0?

Question

find all soluation..............

Answer 1

(sin2x=2sinxcosx)
sin2x-cosx=0
2sinxcosx-cosx=0
cosx(2sinx-1)=0

cosx=0 x=90+360n / x=270+360n

2sinx-1=0
sinx=1/2 x=30+360n / x=150+360n

Answer 2

cosx - (sin2x)(cosx) = 0 reverse distributive property: cosx (1-sin2x) = 0 so that means: cosx = 0 and sin2x = 0 for the cosx on the unit circle, cos pi/2 and cos 3pi/2 equal 0. so: x = pi/2 an x = 3pi/2. for the sin2x, we know that at 0 and pi, the sin is 0. so: sin 2x = 0 and sin2x = pi 2x = 0 and 2x = pi x = 0 and x = pi/2.

Answer 3

sin2x = cosx
2sinxcosx = cosx
2sinxcosx-cosx = 0
cosx(2sinx-1) = 0
cosx = 0 or (2sinx-1) = 0

1. cosx = 0(because your question metioned 0 so, x = 90(degrees) or 270(degrees)
2. 2sinx = 1
sinx = 1/2(because your question metioned 0 so, x = 30(degrees) or 150(degrees)

Therefore, the "whole" answer to your question is
30(degrees) or 90(degrees) or 150(degrees) or 270(degrees).

Answer 4

sin 2x - cos x = 0

We need to use the sine double angle trig identity that sin(2x) = 2 cos(x) sin(x). So...

2 sinx cosx - cosx = 0
==> factor
cosx (2sinx - 1) = 0

So, we have two equations:

cos x = 0 *AND* (2sinx - 1) = 0

First, cos x = 0
==> x = π/2, 3π/2, 5π/2...

Next, 2 sinx - 1 = 0
2 sinx = 1
sinx = 1/2
==> x = π/6, 5π/6, 13π/6...

So, we need to put our answers in terms of any integer 'n,' so we have three sets of answers,

x = π/2 ± πn, π/6 ± 2πn, or 5π/6 ± 2πn, where n is any integer.

Answer 5

2 sinx cos x - cos x = 0
cos x (2 sin x - 1) = 0
cos x = 0 , sin x = 1 / 2
x = 90° , 270° , 30° , 150°

Answer 6

The first answerer was close but still wrong

Here's the double angle formula for sin:
sin(2x) = 2sin(x)cos(x)

So
sin(2x) - cos(x) = 0
2sin(x)cos(x) - cos(x) = 0
cos(x) [2sin(x) - 1] = 0

cos(x) = 0
x = pi/2, 3pi/2, etc.

2sin(x) - 1 = 0
sin(x) = 1/2
x = pi/6, 5pi/6, etc.

So the solution is:
..., pi/6, pi/2, 5pi/6, 3pi/2, ...

Answer 7

sin2x = cos x.

sin double angle formula:
sin2x = 2(sin x)(cos x)

2(sin x)(cos x) = cos x
2sin x = 1
sin x = 1/2
x = pi/6 or 5pi/6

also note, that adding 2npi to either solution is also solution, n being any integer.

Answer 8

x == -[Pi]/2 + 2 [Pi]n
x == [Pi]/2 + 2 [Pi] n
x == [Pi]/6 + 2 [Pi] n
x == (5 [Pi])/6 + 2 [Pi] n

where n are integers.