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Find the general solution of : 2(sinx - cos2x) - sin2x (1+2sinx) +2cosx = 0
Ans:x= 2nΠ or x= nΠ(-1)^n (-Π/2) Can u explain it?
Ans:x= 2nΠ or x= nΠ(-1)^n (-Π/2) Can u explain it?
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Find the general solution of : 2(sinx - cos2x) - sin2x (1+2sinx) +2cos...
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Find the general solution of : 2(sinx - cos2x) - sin2x (1+2sinx) +2cos...
Explanation:
We are given the equation:
2(sinx - cos2x) - sin2x (1 + 2sinx) - 2cosx = 0
Expanding the equation, we get:
2sinx - 2cos2x - sin2x - 2sinx cos2x - 2cosx = 0
Grouping the terms:
(2sinx - sin2x - 2sinx cos2x) - 2cos2x - 2cosx = 0
Factoring out sinx:
2sinx(1 - cos2x - cosx) - 2cos2x - 2cosx = 0
Using the identity sin2x = 2sinx cosx:
2sinx(1 - cos2x - cosx) - 2cos2x - 2cosx = 0
2sinx(1 - cosx(1 + 2sinx)) - 2cos2x - 2cosx = 0
Dividing both sides by 2:
sinx(1 - cosx(1 + 2sinx)) - cos2x - cosx = 0
Solving the equation:
To solve this equation, we need to find the values of x that make it true.
First, let's look at the term inside the brackets, 1 - cosx(1 + 2sinx).
This term is equal to 0 when:
cosx(1 + 2sinx) = 1
1 + 2sinx = secx
2sinx = secx - 1
sinx = (secx - 1)/2
Using the identity secx = 1/cosx:
sinx = (1 - cosx)/2cosx
2sinx = 1 - cosx
Substituting this into the original equation, we get:
(1 - cosx) - cos2x - cosx = 0
1 - 2cosx - cos2x = 0
(1 - cosx)^2 = 0
cosx = 1
x = 2nπ
or
cosx = -1
x = (2n + 1)π/2
Substituting sinx = (1 - cosx)/2 into the original equation, we get:
2(1 - cosx) - sin2x (1 + 2sinx) - 2cosx = 0
2(1 - cosx) - 2sinx(1 - cosx)(1 + 2sinx) - 2
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Find the general solution of : 2(sinx - cos2x) - sin2x (1+2sinx) +2cosx = 0 Ans:x= 2nΠ or x= nΠ(-1)^n (-Π/2) Can u explain it?
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