SOLUTION: Solve cos2x = 1 - sinx for 0° < x < 360°. In terms of angles
Algebra.Com
Question 1180393: Solve cos2x = 1 - sinx for 0° < x < 360°. In terms of angles
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Using the trig identity cos(2x) = 1 - 2sin^2(x), we have:
1 - 2sin^2(x) = 1 - sin(x) -> sin(x)(2sin(x) - 1) = 0
This equation is satisfied if either sin(x) = 0 or 2sin(x) = 1
sin(x) = 0 -> x = 0, pi
sin(x) = 1/2 -> x = pi/6, 5pi/6
RELATED QUESTIONS
solve for 0<=x<=360... (answered by Theo)
Solve 2 sinx + cos2x = 2sin.sinx - 1 for 0 less thAN or equal and x greater than or equal (answered by jsmallt9)
Heres 3 questions I was curious about
Given sin A = 1/3 A in Quad II, CosB = -3/5 B in (answered by drk)
Solve for all values of x if x is measured in degrees.
2cos^2x=3sinx
2(1-sin^2x)=3sinx
(answered by stanbon)
Solve the equation sqr3 sinx = 1/2 sec x for 0 deg < x < 360... (answered by lwsshak3)
Solve for x on the interval given:
a) 2sin^2x-5cosx+1=0 [0, pie]
b) sinx/2=0 [-pie,... (answered by Alan3354)
how do i solve the following equation for values of x in the range
0 degrees < x <360... (answered by Edwin McCravy)
Solve the following equation for x
2cos^2x+sinx=1 for 0 < x < 360 deg
please... (answered by lwsshak3)
Solve for x in [0, 2pi]:... (answered by jsmallt9)