Solving: sin(x) + 1 = 2cos(x)?

Question

how can i approach it besides graphing.
thank you.

SOLVING
therefore, trying to find the value of 'x'

i think willie got it.
thank you!

Answer 1

sin(x) + 1 = 2cos(x) Square both sides and get

sin^2(x) + 2sin(x) + 1 = (sin(x) + 1)^2 = (2cos(x))^2 = 4cos^2(x)

= 4(1 - sin^2(x)) = 4 - 4sin^2(x). Since sin^2(x) + 2sin(x) +1 = 4 - 4sin^2(x) then 5sin^2(x) + 2sin(x) - 3 = 0.

Use the quadratic formula and solve for sin(x).

sin(x) = (-2 +/- squareroot(2^2 - 4(5)(-3))/(2)(5)

= (-2 +/- squareroot(4 + 60))/10

= (-2 +/- squareroot(64))/10 = (-2 +/- 8)/10

sin(x) = (-2 + 8)/10 = 6/10 = 3/5
or sin(x) = (-2 - 8)/10 = -10/10 = -1

If sin(x) = 3/5 then sin^2(x) = 9/25 and cos^2(x) = 1 - sin^2(x)
= 1 - 9/25 = (25 - 9)/ 25 = 16/25,
so cos(x) = +/- 4/5. cos(x) = -4/5 doesn't work and must be eliminated, but if cos(x) = 4/5 then 2cos(x) = 8/5 = 5/5 + 3/5
= 1 + sin(x)
so the angle x such that sin(x) = 3/5 and cos(x) = 4/5 is a solution. (x is about 37 degrees.)

If sin(x) = -1 then x = 270 degrees, and cos(270) = 0.
Since sin(270) + 1 = -1 + 1 = 0 = (2)(0) = 2cos(270),
then 270 degrees is another solution.

Answer 2

r u trying to solve for x? or do u have to prove that sin x + 1 =2cos x? please clarify.