I Need help solving some trigonometric equations.?

Question

1.) sinx= cosx on [0, 2pi]
2.) sin2x = 1/2 on [0, 2pi]
3.) sin3x= sqrt(3)/2 for all x
4.) cos4x= -1/2 for all x
5.) tan2x=0 on [pi/2, 3pi/2]

Answer 1

1)
sin x = cos x
when x = pi/4 and 5pi/4

2)
sin u = 1/2 when u = pi/6, 11pi/6, 13pi/6, and 23pi/6 (because it's 2x, you need to go around the circle twice)

so...
2x = pi/6, 11pi/6, 13pi/6, and 23pi/6
x = pi/12, 11pi/12, 13pi/12, and 23pi/12

3)
sin u = sqrt 3 / 2 when
u = pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3, 17pi/3

So...
3x = pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3, 17pi/3
x = pi/9, 5pi/9, 7pi/9, 11pi/9, 13pi/9, 17pi/9

4)
cos u = -1/2 when
u = 5pi/6, 7pi/6, 17pi/6, 19pi/6, 29pi/6,
31pi/6, 41pi/6, 43pi/6

So,
4x = 5pi/6, 7pi/6, 17pi/6, 19pi/6, 29pi/6,
31pi/6, 41pi/6, 43pi/6

divide all by 4 to get what x =

5)
tan u = 0 when
u = pi and 3pi

so
2x = pi and 3pi
x = pi/2, 3pi/2

Answer 2

This is asking you to remember some of the "easily-memorizable" values of the trig functions.

Sin = cos, only when the reference angle is 45 degrees (pi/4 and also 5pi/4, but not 3pi/4 or 7pi/4)

Sin is 1/2 when the reference angle is 30 degrees (pi/6), but you're given 2x instead of x, so 2x = pi/6... x = pi/12. This also works when the angle is 5pi/6 (150 deg), so 2x=5pi/6, x=5pi/12

sqrt(3)/2 is the sin of 60 degrees (pi/3). Since they gave you 3x, 3x=pi/3, x= pi/9. The other value is 120deg (2pi/3). 3x=2pi/3, x=2pi/9.

cos4x = -1/2 works when 4x is ±2pi/3 and odd-multiples thereof, i.e. ±2n*pi/3 when n is odd. so x = ±n*pi/6 where n is odd

tan2x = 0 only if the angle is 0 or 180 (0 or pi). So 2x=pi, x=pi/2. For the given interval, the only value that works is pi/2.

Answer 3

Lets start
tan x= 1 so x= pi/4 and pi/4+pi
sin2x =1/2 2x = pi/6 +2kpi and x = pi/12+k*pi
so x= pi/12 and 13pi/12.Also 2x=5pi/6+2kpi so x=5pi/12 and17pi/12
3) 3x = pi/3+2kpi so x= pi/9 +2kpi/3 and 3x=2pi/3 +2kpi
x= 2pi/9+2kpi/3

4)4x= 3pi/4+2kpi x= 3pi/16+kpi/2 and 4x= 5pi/4+2kpi
so x= 5pi/16++kpi/2
tan 2x = 0 2x = kpi so x= kpi/2 so x=pi/2 and pi

Answer 4

MEMORIZE!!!

Functions on the
Unit Circle
θ. . . sinθ . . .cosθ
0. . . √(0/4) .√(4/4)
π/6 ,√(1/4) . √(3/4)
π/4. √(2/4) . √(2/4)
π/3. √(3/4) . √(1/4)
π/2. √(4/4) . √(0/4)

sin(x)= cos(x) where?
sin(2x)=1/2 where? Then what must x be?
cos(3x)=√(3/4) = (√3)/2 where? Then x = what?

tan(2x) = sin(2x)/cos(2x) How can you get that to be zero? Hint: sin(2x) has to = 0.Then what must x be?