# The values of x that are solutions to the equation cos^2x=sin 2x in the interval [0,π] are: a. arctan 1/2, 0, and π/2 b. arctan 1/2 and π/2 c. arctan 1/2 and 0 d. arctan 1/2 and π e....

The values of x that are solutions to the equation cos^2x=sin 2x in the interval [0,π] are:

a. arctan 1/2, 0, and π/2

b. arctan 1/2 and π/2

c. arctan 1/2 and 0

d. arctan 1/2 and π

e. arctan 1/2 only

jeew-m | College Teacher | (Level 1) Educator Emeritus

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We know that `sin2x = 2sinxcosx`

`cos^2x = sin2x`

`cos^2x = 2sinxcosx`

`cos^2x-2sinxcosx = 0`

`cosx(cosx-2sinx) = 0`

`cosx = 0` OR `cosx-2sinx = 0`

`cosx = 0`

`cosx = cos(pi/2)`

`x = 2npi+-pi/2` where `n in Z`

`n = 0` then `x = +-pi/2`

`n = 1` then `x = (5pi)/2` OR `x = (3pi)/2`

`cosx-2sinx= 0`

`(sinx)/(cosx) = 1/2`

`tanx = 1/2`

`tanx = tan(arctan1/2)`

`x = mpi+arctan1/2` where `m in Z`

`m = 0` then `x = arctan1/2`

`m = 1` then `x = pi+arctan1/2`

So the answers between (0 and pi) would be;

`x = arctan(1/2)` and `x = pi/2`

So the correct answer is b)