Solve the equation : 2*sin^2 x + cos x - 1 = 0

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The equation `2*sin^2 x + cos x - 1 = 0` has to be solved for x.

This trigonometric equation has both sin x as well as cos x. It is possible to write `sin^2x = 1 - cos^2x` . This eliminates sin x and would allow the equation to...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

The equation `2*sin^2 x + cos x - 1 = 0` has to be solved for x.

This trigonometric equation has both sin x as well as cos x. It is possible to write `sin^2x = 1 - cos^2x` . This eliminates sin x and would allow the equation to be written as a quadratic in terms of cos x.

`2*sin^2 x + cos x - 1 = 0`

=> `2*(1 - cos^2x) + cos x - 1 = 0`

=> `2 - 2*cos^2x + cos x - 1 = 0`

=> `-2*cos^2x + cos x + 1 = 0`

=> `2*cos^2x - cos x - 1 = 0`

=> `2*cos^2x - 2*cos x + cos x - 1 = 0`

=> `2*cos x(cosx - 1) + 1(cos x - 1) = 0`

=> `(2*cos x + 1)(cos x - 1) = 0`

`2*cos x + 1 = 0 => cos x = -1/2`

`cos x - 1 = 0 => cos x = 1`

The cosine function has a periodicity of `2*pi` . This gives an infinite number of solutions of the equation.

`cos x = -1/2`

=> `x = 2*pi/3 + 2*n*pi` and `x = 4*pi/3 + 2*n*pi`

cos x = 1

=> `x = 2*n*pi`

The solutions of the equation `2*sin^2 x + cos x - 1 = 0` are `{2*pi/3 + 2*n*pi, 4*pi/3 + 2*n*pi, 2*n*pi}`

Approved by eNotes Editorial Team