Can you solve this by using trig. identities?

Question

Sin x - 2Cos^2 x + 1 = 0

I can do these but then I get stuck because I dont know what to do with the COSINE

Answer 1

sin x - 2 ( 1 - sin ² x ) + 1 = 0
2 sin ² x + sin x - 1 = 0
( 2 sin x - 1 ) ( sin x + 1 ) = 0
sin x = 1/2 , sin x = - 1
x = 30° , 150° , 270°
OR
x = π/6 , 5π/6, 3π/2

Answer 2

Don't understand the difficulty - am I missing something? This is my knee-jerk reaction:
Sin x -2Cos^2 x +1 = 0

Since Cos^2 x = 1-Sin^2 x ,
Sin x -2(1-Sin^2 x) + 1 = 0
i.e.
2Sin^2 x +Sin x - 1 = 0, an easy quadratic in Sin x that has the solutions
Sin x = -1 (Cos x = 0) and
Sin x = 1/2 (Cos x = Sqrt(3)/2)

Answer 3

Use sin^2(x) + cos^2(x) = 1 to get cosine in terms of sine. Then let p = sin(x), rewrite in terms of p, and solve for p. (You'll end up with a quadratic equation in terms of p which is easy to factor.) Then use that to find x = sin^-1(p).

Answer 4

Sin x - 2Cos^2 x + 1 = 0
Sin x -Cos^2 x -cos^2 x+ 1 =0
sin x -cos ^2x+sin ^2x =0 (as 1-cos^2 x = Sin ^2 x)
sinx -cos 2x =0 (as cos ^2x - sin^2 x = cos 2x)

Answer 5

use cos^2 x = 1 - sin^2 x
then solve the quadratic in sinx

Answer 6

sin x - 2cos^2 x + 1 = 0
=> sin x - ( 2cos^2 x - 1) = 0
=> sin x - cos 2x = 0
=> cos ( π/2 - x) = cos 2x
Using the formula, cos α = cos β => α = 2kπ ± β
=> 2x = 2kπ + π/2 - x or 2x = 2kπ - π/2 + x
=> x = 2kπ/3 + π/6 or x = 2kπ - π/2, k belongs to Z.

Answer 7

sin(x)-2(1-sin²x)+1=0
2sin²(x)+sin(x)-1=0
(2sin(x)-1)(sin(x)+1)=0
sin(x)=1/2 and sin(x)=-1
x=π/6,5π/6 and x=3π/2

Note: I only gave positive values between 0 and 2π. There are, of course, more values with each rotation and when using negative values

Answer 8

sinx - cos2x = 0
cos2x = sinx = cos(pi/2 - x + 2n*pi)
2x = pi/2 - x + 2n*pi
Solve for x,
x = pi/6 + 2n*pi/3, n can be any integer
Solution:
pi/6, 5pi/6, 9pi/6, ...
or
cos2x = sinx = cos(-pi/2 + x + 2n*pi)
2x = -pi/2 + x + 2n*pi
Solve for x,
x = -pi/2 + 2npi, n can be any integer
Solution:
3pi/2, 7pi/2, ...

Answer 9

(csc x - cot x)^2 = (1-cos x)/(1+cos x) (csc x - cot x)^2 = 1-cos^2(x) (csc x - cot x)^2 = sin^2(x) csc x - cot x = sin(x) 1 - cosx = sin^2(x)

Answer 10

yes
use cos double angle identities