Solve for all real numbers cos 2x +sin x=0.

1 Answer | Add Yours

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Given `cos 2x +sin x=0`

`rArr 1-2sin^2x+sinx=0` `(cos2x=cos^2x-sin^2x=1-2sin^2x)`

Let, `t = sinx` , then the above equation reduces to the quadratic form `-2t^2+t+1=0`

The solutions of which can be obtained from the quadratic formula such that:

`t=(-1+-sqrt(1^2-4*(-2)*1))/(2*(-2))`

`rArr t=-1/2 and t=1`

Replacing `t` with `sinx` , we have,

`sinx = -1/2 ` and `sinx = 1` .

For `sinx=-1/2` , notice that sinx is negative in the IIIrd and IVth quadrant,

Therefore, when `sinx=-1/2, (x=pi+pi/6)` (IIIrd quardant) and `(x=2pi-pi/6)` (IVth quardant)

i.e. `x=(7pi)/6 and (11pi)/6` .

In general `x = kpi+(-1)^(k+1)*pi/6` where, k is an integer.

Again for `sinx=1` ,

i.e. `x=pi/2, (5pi)/2` etc.

In general `x = kpi+(-1)^k * pi/2` where, k is an integer.

Thus, all solutions are given by `x = kpi+(-1)^(k+1) * pi/6 and kpi+(-1)^k * pi/2` where, k is an integer.

Sources:

We’ve answered 318,960 questions. We can answer yours, too.

Ask a question