Find all solutions in the interval [0, 2π) sin2x + sin x = 0

1 answer:

3 0

Sin 2 x + sin x = 0
( sin 2 x = 2 sin x cos x )
2 sin x cos x + sin x = 0
sin x ( 2 cos x + 1 ) = 0
sin x = 0
x 1 = 0
x 2 = π
or: 2 cos x + 1 = 0
2 cos x = - 1
cos x = -1/2
x 3 = 2π/3
x 4 = 4π/3

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Solve the system of equations. x+y=4 y=x^2 - 8x + 16 a) {(-3,7).(-4, 8)} b) [(4,0)} c) {(3,1),(4,0) d) {(3,7). (4.0)} e) none

Marysya12 [62]

Answer: The required solution of the given system is

(x, y) = (3, 1) and (4, 0).

Step-by-step explanation: We are given to solve the following system of equations :

$x+y=4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y=x^2-8x+16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

From equation (i), we have

$x+y=4\\\\\Rightarrow y=4-x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of y from equation (iii) in equation (ii), we get

$y=x^2-8x+16\\\\\Rightarrow 4-x=x^2-8x+16\\\\\Rightarrow x^2-8x+16-4+x=0\\\\\Rightarrow x^2-7x+12=0\\\\\Rightarrow x^2-4x-3x+12=0\\\\\Rightarrow x(x-4)-3(x-4)=0\\\\\Rightarrow (x-3)(x-4)=0\\\\\Rightarrow x-3=0,~~~~~~~x-4=0\\\\\Rightarrow x=3,~4.$