2sin^2x - cosx = 1?

Question

How do I find all the solutions in between [0, 2pi)??

Answer 1

For this question, the trigonometry identity (sin x)^2 + (cos x)^2 = 1 is used to convert the original equation into a quadratic equation.


We make (sin x) the subject from the identity
(sin x)^2 + (cos x)^2 = 1,
(sin x)^2 = 1 - (cos x)^2

Multiply both sides by 2, we obtain
2(sin x)^2 = 2 - 2(cos x)^2


Substitute 2(sin x)^2 = 2 - 2(cos x)^2 into the original equation 2(sin x)^2 - cos x = 1,
We obtain
2 - 2(cos x)^2 - cos x = 1

Rearrange the equation so that we have zero on one side of the equation,
2(cos x)^2 + cos x - 1 = 0


We replace cos x with P,
The quadratic equation is 2P^2 + P - 1 = 0

We factorise the quadratic equation,
(2P - 1) (P + 1) = 0

By changing P back into cos x,
(2cos x - 1) (cos x + 1) = 0

we have these two equations to solve since
2cos x -1 = 0 or cos x + 1 = 0


For the first equation,
2cos x - 1 = 0
2cos x = 1
cos x = 1/2
Since cosine values are positive in the first and fourth quadrant, possible values of x = pi/3 (first quadrant), 5 pi/3 (fourth quadrant)

For the second equation,
cos x + 1 = 0
cos x = -1
x = pi

The solutions of x are pi/3 , pi , 5 pi/3

I hope this helps

Answer 2

sin^2 (x) = 1 - cos^2 (x)
substitute
2[1- cos^2 (x) ] - cos x =1
simplify
2 -2 cos^2 (x) - cos (x) - 1 = 0
simplify again
-2 cos^2 (x) - cos (x) + 1 = 0
multiply by -1
2cos^2 (x) + cos (x) - 1 = 0
either factor [as if an algebraic expression] or use the quad formula...

OK?

Answer 3

sin^2x+cos^2x=1
so if you substitue in your equation you get
2(1-cos^2x)-cosx-1=0
or -2cos^2x-cosx+1=0

Now, let y=cosx and solve