Question

Question #9d454

Answer 1

See below.

Explanation:

`sin x + cos x + 2.sqrt(2) . sin x . cos x= 0` or

`cosx = -sinx/(1+2sqrt2 sinx)`

now making `y = sinx` we have

`sqrt(1-y^2) = -y/(1+2sqrt2 y)^2` or squaring both sides

`1-y^2=y^2/(1+2sqrt2y)^2` or

`(1-y^2)(1+2sqrt2y)^2-y^2=0` or

`(2y^2+2sqrt2y+1)(-4y^2+2sqrt2y+1)=0` then we have

`{(2y^2+2sqrt2y+1=0->{(y=-1/sqrt2),(y=-1/sqrt2):}),(-4y^2+2sqrt2y+1=0->{(y=1/4 (sqrt[2] - sqrt[6])),(y=1/4 (sqrt[2] + sqrt[6])):}):}`

or

`sinx={(-1/sqrt2->x = (5pi)/4 + 2 k pi),(1/4 (sqrt[2] - sqrt[6])->x = arcsin(1/4 (sqrt[2] - sqrt[6]))+2kpi),(1/4 (sqrt[2] + sqrt[6])->x = pi-arcsin(1/4 (sqrt[2] + sqrt[6]))+2kpi):}`

Answer 2

`x = (5pi)/4 + 2kpi`
`x = (7pi)/12 + 2kpi`
`x = (23pi)/12 + 2kpi`

Explanation:

`sin x + cos x + 2sqrt2.sin x.cos x = 0` (1)
Call sin x + cos x = t
Square both sides:
`(sin x + cos x)^2 = sin^2 x + cos ^2 x + 2cos x.sin x =`
`= 1 + 2cos x.sin x = t^2` -->
`2cos x.sin x = (t^2 - 1)`
`2sqrt2.cos x.sin x = sqrt2(t^2 - 1)`
The equation (1) becomes:
`t + sqrt2(t^2 - 1) = 0`
`sqrt2t^2 + t - sqrt2 = 0`
Solve this quadratic equation for t
`D = d^2 = b^2 - 4ac = 1 + 8 = 9` --> `d = +- 3`
There are 2 real roots:
`t = - b/(2a) +- d/(2a) = - 1/(2sqrt2) +- 3/(2sqrt2) =`
`t1 = - 4/(2sqrt2) = - 2/sqrt2 = - sqrt2`
`t2 = 2/(2sqrt2) = sqrt2/2`
a. `sin x + cos x = t1 = - sqrt2`
Use trig identity: `sin a + cos a = sqrt2.cos (a - pi/4)`
`cos (x - pi/4) = - sqrt2/sqrt2 = - 1` -->
`(x - pi/4) = pi` --> `x = pi + pi/4 = (5pi)/4`,
b. `sin x + cos x = t2 = sqrt2.cos (x - pi/4) = sqrt2/2`
`cos (x - pi/4) = 1/2`
`x - pi/4 = pi/3` --> `x = (pi)/3 + pi/4 = (7pi)/12`
`x - pi/4 = (5pi)/3` --> `x = (5pi)/3 + pi/4 = (23pi)/12`