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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:
`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.
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