solve algebreically the equation
cos2x=1-sinx
also cosx=cosx-1
and please also sin2x=3sinx
Isn't cos2x=cos^2x-sin^2x=1-2sin^2 x ?
put that into the left side, gather terms, and you have a quadratic equation.
The second makes no sense.
To solve the equation cos2x = 1 - sinx algebraically, we can use the double-angle formula for cosine, which states that cos2x = cos^2x - sin^2x.
So, substituting cos^2x - sin^2x for cos2x in the equation, we have:
cos^2x - sin^2x = 1 - sinx
Rearranging the equation, we get:
cos^2x - sin^2x + sinx - 1 = 0
Using the trigonometric identity cos^2x = 1 - sin^2x, we can simplify further:
(1 - sin^2x) - sin^2x + sinx - 1 = 0
Expanding and rearranging terms:
1 - 2sin^2x + sinx - 1 = 0
-2sin^2x + sinx = 0
Factoring out sinx:
sinx(-2sinx + 1) = 0
Now, set each factor equal to zero and solve for x:
sinx = 0 --> x = 0, π, 2π, ...
-2sinx + 1 = 0 --> sinx = 1/2 --> x = π/6 + 2πn, 5π/6 + 2πn
where n is an integer.
For the equation cosx = cosx - 1, notice that both sides of the equation are equal, which means any value of x satisfies this equation. So, there is no specific solution for this equation.
Lastly, for the equation sin2x = 3sinx, we can use the double-angle formula for sine, which states that sin2x = 2sinxcosx.
Substituting 2sinxcosx for sin2x in the equation, we have:
2sinxcosx = 3sinx
Rearranging the equation, we get:
2sinxcosx - 3sinx = 0
Factoring out sinx:
sinx(2cosx - 3) = 0
Now, set each factor equal to zero and solve for x:
sinx = 0 --> x = 0, π, 2π, ...
2cosx - 3 = 0 --> cosx = 3/2
However, there is no solution for cosx = 3/2 since the cosine function only ranges from -1 to 1. Therefore, sin2x = 3sinx has no solution.