Solving trigonometric equations, grade 11 stuff!?

Question

How would I go on about solving:

1. Solve each equation for 0 <= x <= 360
a) cosX = 0
b) 2costx-3=0
c) tanx = sqrt(3)

a) cos^2x-1=sin^2x .. sine square x
b) cos^2x+4=4cosX
c)2sin^2x-7sinx=4

Thanks for any help, or even all :)

Answer 1

DONT GET SCARED BY THE SIZE OF THE ANSWER I HAVE TYPED BELOW---I AM TRYING TO EXPLAIN THE FUNDAMENTALS AND HOW TO REMEMBER STUFF EASILY DURING AN EXAM!!!
0 If you remember the unit circle, and chose any random point on the circumference of the unit circle, and let's suppose the points are {1/2, sq.(3)/2}. Then the x-co-ordinate is the cosine value and the Y co-ordinate is the sine value.
Therefore, cos(60)=1/2 and sin(60)=sq.(3)/2 [sq.=square root]
TRIAL 1: At 0 dergrees, your point would be at
X co-ord.= cos(0)=1 and
Y co-ord.= sin(0)=0
So, the point would be at [1,0]
Draw a unit circle on the corner of your paper during an exam, it helps.
So, 0 degrees or X=0 is not the answer!
TRIAL 2: At 90 degrees, your point would be at
X co-ord.= cos(90)=0
Y co-ord.= sin(90)=1
This works!!! cos(90) 0!
Therefore, X=90

Instead of doing all this mayhem, you could just remember the sine values of 0, 30, 45, 60 and 90 (and remembering 180 wouldn't hurt!). If you know these by-heart, then, you could just use the Pythadorean Theorem to solve for cosine!
X^2 + Y^2 =1 (where X= cosine and Y= sine)---> equation(1)

Suppose, you knew the sine value for 90 before you asked this question, then you could have:
X^2 + (sin(90))^2 =1 (since you have an idea of the unit circle, you substitued 90 for X as you know sin(90)=1 because
sine and cosine values are just opposite in direction)
sin(0)=0 cos(0)=1
sin(30)=1/2 cos(30)=sq.(3)/2
sin(45)= sq.(2)/2 cos(45)=sq.(2)/2
sin(60)=sq.(3)/2 cos(60)=1/2
sin (90)=1 cos(90)=0
{Notice that the values of sine and cosine are in palindromic sequence}

Now, going back to equation (1),
X^2 + (1)^2 =1 (since sine(90)=1)
X^2 = 1-1
X^2 = 0
=> X = 0 when sine was at 90 degrees, implying that cosine (or X) is 0 at 90 degrees!

Now tangent is just the value of sine divided by cosine
So, sin(60)/cos(60) = tan(60)
=> sq.(3)/2 divided by 1/2
would equal tan(60)
i.e sq.(3)

For solving others, you need to remember the 3 commandments of Trigonometry:

sin^2 (x) + cos^2 (x) = 1
tan^2 (x) + 1 = sec^2 (x)
1 + cot^2 (x) = csc^2 (x)

Answer 2

Question1a
x = 90°, 270°
Question1b
Assume question should read as:-
2cosx - √3 = 0
cos x = √3/2
x = 30° , 330° (1st and 4th quadrants)
Question 1c.
x = 60° , 240° (1st and 3rd quadrants)
Question 2a
1 - 2sin²x - 1 = sin²x
3sin²x = 0
sin x = 0
x = 0°,180°
Question 2b
2cos²x - 1 + 4 = 4.cosx
2cos²x - 4cosx + 3 = 0
cosx = [ 4 ± √(16 - 24)] /4
cos x = [4 ±√(-8)] /2
This looks dodgy as we are now into complex numbers!
May be that question is incorrect?
Question 2c
2sin²x. - 7sinx - 4 = 0
(2.sinx + 1).(sinx - 4) = 0
sinx = -1/2
x= 210°, 330°