NEED MATH EXPERT please solve problem with steps written out?

Question

Find the exact solutions of:
(2)^(1/2)cos(x+(pie/4))-1=0 For x between (0,4pie)

By the way the answers have to be in terms of pie for example one example is (3pie/2)

PLEASE NO CALC, EASY WAY FOR 11 GRADE

Answer 1

cos(x+pi/4)=1/sqrt2
a=x+pi/4=basic angle=pi/4
alter the range:
0-4pi---->pi/4-17pi/4
x+pi/4=pi/4, 2pi-pi/4,2pi+pi/4,4pi-pi/4,4pi+pi/4
=pi/4,7pi/4,9pi/4,15pi/4,17pi/4
therefore x= ans-pi/4
= 0,3pi/2,2pi,7pi/2,4pi

Answer 2

Calling the exponent z

2^z =1 implies that the exponent is zero

so cos(x+pi/4)=0 so x+pi/4 = pi/2 + 2npi or x+pi/4=3pi/2+2npi

As you are looking for solutions between 0,4pi

you get x=pi/4 , x=pi/4 +2pi , x = 5pi/4 and x=5pi/4 +2pi

In order x=pi/4, x= 5pi/4, x=9pi/4,x=13pi/4

Answer 3

I hope I've interpreted this one correctly.

If (2)^(1/2)cos(x+(pi/4))-1=0 then
(2)^(1/2)cos(x+(pi/4)) = 1.
Take log (base 2) of both sides, log 1 = 0 in all bases, so
(1/2)cos(x+(pi/4))=0
so cos(x + pi/4) = 0
Let y = x + pi/4 (means x = y - pi/4)
This problem now becomes, cos y = 0, so y = pi/2, 3pi/2, 5 pi/2, 7pi/2.
Hence x = pi/4, 5pi/4, 9pi/4, 13pi/4.

Answer 4

2^(1/2) * cos(x+pi/4) = 1

cos (x + pi/4) = 1/(2^(1/2))

Cos(x+pi/4)=pi/4 or 7*pi/4, so x = 0 or x = 3*pi/2

If the range is 0,4*pi, then adding 2*pi will be also a solution, and we have also 2*pi and 7*pi/2.

It is pi or Pi, not pie.

Answer 5

2[3x+6]= -[x+9] Given 6x + 12 = -x - 9 Distributive resources 6x + 12 - 12 = -x - 9 - 12 Subtract 12 from both area 6x = -x - 21 Simplify 6x + x = -x + x - 21 upload x from both area 7x = -21 Smplify 7x/7 = -21/7 Divide both area through 7 x = -3 Simplify

Answer 6

x = 0, 2 * pi, 4 * pi

Answer 7

cos (x+ pi/4) = 1/(2^(1/2))
x+ pi/4 = 2*n*pi +- pi/4
ie the general sol.
put integral n
n=0 take +ve x= 0
n=1 take +ve x= 2.5 pi
n=1 take -ve x= 1.5 pi
so on...................................