The general solution of 2 cos²x-5sinx-4=0 is x = 7π / 6 or x = 11π / 6 considering [0 < x < 2π]
Given: To find general solution of 2cos²x - 5sinx - 4 = 0.
Let's solve the given equation.
2cos²x - 5sinx - 4 = 0
We know sin²x + cos²x = 1
cos²x = 1 - sin²x
Replacing cos²x in the equation 2cos²x - 5sinx - 4 = 0, we get:
=> 2(1 - sin²x) - 5sinx - 4 = 0
=> 2 - 2sin²x - 5sinx - 4 = 0
=> - 2sin²x - 5sinx + (2 - 4) = 0
=> - 2sin²x - 5sinx + (-2) = 0
=> - 2sin²x - 5sinx - 2 = 0
=> - (2sin²x + 5sinx + 2) = 0
∴ 2sin²x + 5sinx + 2 = 0
Now factorizing 2sin²x + 5sinx + 2 = 0, we get:
2sin²x + 5sinx + 2 = 0
=> 2sin²x + 4sinx + sinx + 2 = 0
=> 2sinx(sinx + 2) + 1(sinx + 2) = 0
Taking (sinx + 2) common, we get:
=> (sinx + 2)(2sinx + 1) = 0
Now either (sinx + 2) = 0 or (2sinx + 1) = 0
=> sinx = -2 or sinx = -1 / 2
We know that the range of sinx lies between -1 and 1 which is -1 ≤ sinx ≤ 1
or sinx ∈ [-1, 1]
So, for sinx = -2, -2 does not fall in the range [-1, 1] that is -2 ∉ [-1, 1]
Therefore we will not consider sinx = -2.
Now sinx = -1 / 2, and -1 /2 ∈ [-1, 1], so this equation is considered.
NOTE : As no information is given, we assume that x lies in the interval 0 and 2π, that is x ∈ [0, 2π].
We see that sinx = -1 / 2, that is sinx is negative. Now sinx is negative in the IIIrd and IVth Quadrant.
Also there is a formula for finding the value of x that lies in the given domain [0, 2π].
FORMULA: if sinx = - siny
then, sinx = sin(-y) [- sinx = sin(-x) ]
sinx = sin(π + y) or sinx = sin(2π - y) [as x will be in either IIIrd or IVth Quadrant]
So we have: sinx = -1 / 2
=> sinx = -sin(π / 6) [sin (π / 6) = 1 / 2] [0 < x < 2π]
=> sinx = sin(-π / 6) [-sinx = sin(-x)] [0 < x < 2π]
=> sinx = sin(π + π / 6) or sinx = sin(2π - π / 6) [0 < x < 2π]
=> sinx = sin(7π / 6) or sinx = sin(11π / 6) [0 < x < 2π]
Therefore x = 7π / 6 or x = 11π / 6 [0 < x < 2π]
Hence the general solution of 2 cos²x-5sinx-4=0 is x = 7π / 6 or x = 11π / 6 considering [0 < x < 2π]
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