Question

How do you solve for x in `sin2x+cosx=0`?

Answer 1

So `x=pi/2,(7pi)/6`

Explanation:

We know that `sin2x=2sinxcosx`

So this becomes,

`2sinxcosx+cosx=0`

`(2sinx+1)(cosx)=0`

So either `2sinx+1=0` so `sinx=-1/2` in which case `x=(7pi)/6`
Or, `cosx=0` in which case `x=pi/2`

So `x=pi/2,(7pi)/6`

Answer 2

`x=210 or -30 or 330`

Explanation:

`Sin(2x)=2sin(x)cos(x)`

`2sin(x)cos(x)=-cos(x)`

`sin(x)=-1/2`

`sin^-1(x)=x`

sin of x is negative value Only at the third and the forth quarter.

`x=210 or -30 or 330`

Answer 3

`pi/2; (7pi)/6; (3pi)/2; (11pi)/6` for interval `(0, 2pi)`

Explanation:

sin 2x + cos x = 0
2sin x.cos x + cos x = 0
cos x(2sin x + 1) = 0
either factor should be zero.
a. cos x = 0
Unit circle gives 2 solutions -->
`x = pi/2 + 2kpi`, and
`x = (3pi)/2 + 2kpi`.
b. 2sin x + 1 = 0 --> `sin x= - 1/2`
Trig table and unit circle give 2 solutions:
`x = - pi/6 + 2kpi`, or `x = (11pi)/6 + 2kpi` (co-terminal)
`x = pi - (-pi/6) = (7pi)/6 + 2kpi`