Trigonometric equation (URGENT!!)

Question

a)sin(5x)-cos(5x)=0 

b)sen(x)+cos(x)=1 

c)sqrt2*sin(x)/2+ sqrt2*cos(x)/2=-1/2 

pls its a hurry and pls explain step by step. 

pls use cos,sin and tan if needed.

 0 following

Answer 1

a) sin(5x) = cos(5x), divide by cos(x)...

1=sin(5x)/cos(5x) = tan(5x); so what values of x give tan(5x) = 1?

If it was tan(x)=1, that is a 45º (π/4) rt. triangle with opposite and adjacent sides of 1.

It would then be x = +/-n*π/4, where n is an integer 1,2,3 etc.

So for 5x, x=9 to be 45º or π/4, and we have 5x = +/- n*π/4, so x = +/- n*π/20. n is an integer 1,2,3 etc

b) sin(x) + cos (x) = 1, square both sides...

sin²(x) + cos²(x) + 2sin(x)cos(x) = 1, and with sin²(x) + cos²(x) = 1, we get 2sin(x)cos(x) = 0

we know that sin(2x)=2sin(x)cos(x), so we get sin(2x)=0. Again, compare to sin(x) = 0. 2x = +/- nπ for that to be zero, so x = +/- nπ/2, where n is an integer 1, 2, 3, etc.

c) rewriting it clearly,

(sqrt(2)/2)*sin(x)+ (sqrt(2)/2)*cos(x)=-1/2,

sin(x) + cos(x) = -1/sqrt(2)

square both sides...

sin²(x) + cos²(x) + 2sin(x)cos(x) = 1/2, with sin²(x) + cos²(x) = 1 we have,

2sin(x)cos(x) = -1/2, and like the one previously...

sin(2x) = -1/2, so sin(x) = -1/2, and sin(-x) = 1/2. The π/6 rt. triangle with opposite side 1 and hypotenuse 2.

so 2x= +/- nπ/6, x = +/- nπ/12, where n is an integer 1, 2, 3, etc.