Solve: 2sin2x + cos2x = 1?

Question

This is a UK maths C4 question

Answer 1

Let's use "double angle" identities to restate the above:

1)
2sin2x + cos2x = 1
2*2sin(x)cos(x) + 1 - 2sin(x)^2 = 1
4sin(x)cos(x) - 2sin(x)^2 = 0
2sin(x)[2cos(x) - sin(x)] = 0

So we can break this up into a system of questions:

{2sin(x) = 0
{2cos(x) - sin(x) = 0
---
{sin(x) = 0
{2cos(x) = sin(x)
---
{sin(x) = 0
{2 = tan(x)

Our first equation is trivial to solve, as sin(x) is equal to 0 for any x of the form n(pi), where n is any integer.

For our second equation, x will equal the arctan(2) + 2n(pi), where n is any integer.

--charlie

Answer 2

I'm assuming that sin2x means sin^2(x).

2sin^2(x)+cos^2(x)=1
sin^2(x)+sin^2(x)+cos^2(x)=1
Use the trig identity:sin^2(x)+cos^2(x)=1
sin^2(x)+1=1
sin^2(x)=0
sin(x)=0
x=n*PI (in radians), where n is an integer.
x=n*180 (in degrees), where n is an integer.

Answer 3

2sin2x=1-cos2x
2sin2x=sin^2 x+cos^2x-(cos^2x-sin^2x)

2sin2x=2sin^2 x

sin2x=sin^2 x

2sinxcosx=sinx sinx

2sinxcosx-sinx sinx=0
sinx(2cosx-sinx)=0
sinx = 0
x=(2k+1)pi k= 0,1,2,...
or
2cosx=sinx
or tanx=2
x=arc tan x

Answer 4

x = 63.435 degrees