Given that 2cos^2 x - 3sin x -3 = 0, show that 2sin^2 x + 3sin x +1 = 0?
Hence solve 2cos^2 x + 4sin x - 3 = 0, giving all solutions in the interval 0 ≤ x ≤ 2π
2 cos^2 x - 3 sin x - 3 = 0
but
cos^2 x = 1 - sin^2 x
so
2 (1 -sin^2 x) - 3 sin 3 - 3 = 0
2 - 2 sin^2 x - 3 sin x - 3 = 0
2 sin^2 x + 3 sin x + 1 = 0
(2 sin x + 1)(sin x + 1) = 0
sin x = -1/2 or sin x = -1
x = 210 degrees or x = 270 degrees
now do the second half :)
oh, forgot quadrant 4
360 - 30 = 330 degrees
To show that 2sin^2 x + 3sin x + 1 = 0, we need to re-arrange the given equation 2cos^2 x - 3sin x - 3 = 0 by using some trigonometric identities.
First, let's express cos^2 x in terms of sin x using the identity sin^2 x + cos^2 x = 1. Rearranging this identity, we get cos^2 x = 1 - sin^2 x.
Substituting that into the given equation, we have:
2(1 - sin^2 x) - 3sin x - 3 = 0
Simplifying, we get:
2 - 2sin^2 x - 3sin x - 3 = 0
Rearranging terms, we have:
-2sin^2 x - 3sin x - 1 = 0
Divide through by -1 to make the coefficient of the leading term positive:
2sin^2 x + 3sin x + 1 = 0
Hence, we have shown that 2sin^2 x + 3sin x + 1 = 0.
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Now, let's solve the equation 2cos^2 x + 4sin x - 3 = 0 in the interval 0 ≤ x ≤ 2π.
We can start by rearranging the equation as follows:
2(1 - sin^2 x) + 4sin x - 3 = 0
Expanding the brackets, we get:
2 - 2sin^2 x + 4sin x - 3 = 0
Simplifying:
-2sin^2 x + 4sin x - 1 = 0
Divide through by -1:
2sin^2 x - 4sin x + 1 = 0
Now, we can solve this quadratic equation for sin x by factoring or using the quadratic formula.
The factored form would be:
(2sin x - 1)(sin x - 1) = 0
From this, we can find two possible values for sin x:
1) 2sin x - 1 = 0
2sin x = 1
sin x = 1/2
x = π/6 or x = 5π/6
2) sin x - 1 = 0
sin x = 1
x = π/2
Therefore, the solutions in the interval 0 ≤ x ≤ 2π are x = π/6, x = 5π/6, and x = π/2.