cos2x-sinx=0. Factor and solve for sinx
1 - 2sin^2 x - sinx = 0
2sin^2 x + sinx - 1 = 0
(2sinx - 1)(sinx + 1) = 0
sinx = 1/2 or sinx = -1
in degrees, x = 30° , 150° , 270°
in radians, x = π/5 , 5π/6 , 3π/2
Thanks. I was wondering, why is it "+ sin x" in the second step?
do you just change all the signs of the terms? if so, why is this necessary?
I multiplied each term by -1
(notice all the signs have switched)
I like my quadratics to start with a positive coefficient, although that would not be necessary.
It just makes it easier to factor.
ok thank you.
To solve the equation cos(2x) - sin(x) = 0 and find the values of sin(x), we can use algebraic manipulation and trigonometric identities.
Step 1: Recall the double angle identity for cosine:
cos(2x) = 1 - 2sin²(x)
Step 2: Substitute the double angle identity into the equation:
1 - 2sin²(x) - sin(x) = 0
Step 3: Rearrange the equation:
-2sin²(x) - sin(x) + 1 = 0
Step 4: Factor the quadratic equation:
(-2sin(x) + 1)(sin(x) + 1) = 0
Step 5: Set each factor equal to zero and solve for sin(x):
-2sin(x) + 1 = 0 OR sin(x) + 1 = 0
For -2sin(x) + 1 = 0:
2sin(x) = 1
sin(x) = 1/2
For sin(x) + 1 = 0:
sin(x) = -1
Therefore, the solutions for sin(x) are sin(x) = 1/2 and sin(x) = -1.