Question

Find the general solution of equation
sinx + cosx = 2 (root2) (sinx)(cosx)

Answer 1

Note that
sinx + cosx = √2 (1/√2 sinx + 1/√2 cosx) = √2 sin(x + π/4)
so now you just have
√2 sin(x + π/4) = √2 sin2x
and just finish it off now.

Answer 2

I have solve till here
Not getting after this
Plse help

Answer 3

sinx + cosx = 2√2(sinx)(cosx)
square both sides:
sin^2 x + cos^2 x + 2sinxcosx = 8sin^2 x cos^2 x
8sin^2 x cos^2 x - 2sinxcosx - 1 = 0
let sinx cosx = u
8u^2 - 2u - 1 = 0
(4u + 1)(2u -1) = 0
u = -1/4 or u = 1/2

so sinxcosx = -1/4
(1/2)sin 2x = -1/4
sin2x = -1/2
2x = 210° or 2x = 330°
x = 105° or x = 165° <-----> x = 7π/12 or x = 11π/12
but the period of sin2x is 180°, so we have x = 105+180 = 285°, 165+180 = 345°

or
sinxcosx = 1/2
sin 2x = 1
2x = 90°
x = 45° <-----> x = π/4

since we squared, all answers must be verified, using my calculator ....
x = 45° , 165°, 285° <-----> π/4, 11π/12, 19π/12

general solution in terms of radians:
x = π/4 + kπ
x = 11π/12+ kπ
x = 19π/12 + kπ

Answer 4

well, geez. Now you have
x + π/4 = 2x
x = π/4
That takes care of QI, but there are other solutions, in QIII and QIV.
Draw the graphs and you will see two other solutions:
x = 11π/12 and 19π/12