Question

Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π)

Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π)

@Mathematics

Answer 1

hey it is 2sin^2(x)-cox=1 in the interval [0,2pi)

Answer 2

Good - might have been a bit more difficult.

Since sin^2x+cos^2x=1 we have sin^2x=1-cos^2x. Using this in our equation gives:

2(1-cos^2x)-cosx=1.

Multiplying out the LHS and subtracting 1 over gives:

2-2cos^2x-cosx-1=0

Or, rearranging terms:

-2cos^2x-cosx+1=0.

Factoring gives:

(-2cosx+1)(cosx+1)=0.

This gives two equations:

-2cosx+1=0 => cosx=1/2 => x=pi/4 and 7pi/4 in [0, 2pi) and
cosx+1=0 => cosx=-1 .=> x=pi in [0, 2pi).

Solutions: x=pi/4, pi, and 7pi/4 in [0, 2pi).

Answer 3

what about this one
Find all solutions of the equation sec^(2)x−2=0.
The answer is A+Bkπ where k is any integer and 0<A<π/2

Answer 4

find A and B