Find the general solution of equation
sinx + cosx = 2 (root2) (sinx)(cosx)
I have solve till here
Not getting after this
Plse help
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
To find the general solution to the equation sin(x) + cos(x) = 2√2sin(x)cos(x), we must rearrange the equation and apply trigonometric identities. Here's how you can do it:
Step 1: Start by moving all terms to one side of the equation to get sin(x) + cos(x) - 2√2sin(x)cos(x) = 0.
Step 2: Notice that the left side of the equation resembles the double-angle formula for sin(2x): sin(x) + cos(x) = √2sin(x + π/4). Rewriting the equation, we have √2sin(x + π/4) - 2√2sin(x)cos(x) = 0.
Step 3: Factor out √2sin(x) from both terms: √2sin(x)[cos(x) - √2sin(x + π/4)] = 0.
Step 4: Now we have two possibilities for the solution:
i) Set √2sin(x) = 0:
Solving this equation gives sin(x) = 0. The solutions for sin(x) = 0 occur when x = nπ, where n is an integer.
ii) Set cos(x) - √2sin(x + π/4) = 0:
Rearranging this equation gives cos(x) = √2sin(x + π/4).
By squaring both sides, we get cos^2(x) = (√2sin(x + π/4))^2, which simplifies to cos^2(x) = 2sin^2(x + π/4).
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can substitute cos^2(x) = 1 - sin^2(x), giving 1 - sin^2(x) = 2sin^2(x + π/4).
Expanding sin^2(x + π/4) using the double-angle formula, we have 1 - sin^2(x) = 2[sin^2(x)cos^2(π/4) + cos^2(x)sin^2(π/4) + 2sin(x)cos(x)sin(π/4)cos(π/4)].
Simplifying further, we get 1 - sin^2(x) = 2sin^2(x)[1/2 + 1/2 + (1/2)(√2)(√2)].
This reduces to 1 - sin^2(x) = 2sin^2(x)(1 + 1 + 2) = 8sin^2(x).
Rearranging the equation, we have 8sin^2(x) + sin^2(x) - 1 = 0.
Factoring out sin^2(x), we get (8 + 1)sin^2(x) - 1 = 0.
Simplifying further, 9sin^2(x) - 1 = 0.
Solving this quadratic equation for sin^2(x) gives sin^2(x) = 1/9.
Taking the square root of both sides, we have sin(x) = ±1/3.
The solutions for sin(x) = 1/3 occur when x = arcsin(1/3) + 2πn or x = π - arcsin(1/3) + 2πn, where n is an integer.
Similarly, the solutions for sin(x) = -1/3 occur when x = arcsin(-1/3) + 2πn or x = π - arcsin(-1/3) + 2πn, where n is an integer.
Therefore, the general solution is:
x = nπ, arcsin(1/3) + 2πn, π - arcsin(1/3) + 2πn, arcsin(-1/3) + 2πn, π - arcsin(-1/3) + 2πn, where n is an integer.
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.
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