Question 5010: I want to find all solutions of this equation.
2 sin x + square root of 3 = 0
I want the steps to solve over the inteval [0,2pi)
1. sin theta = square root of 3 over 2
2. sin 3x = neg. 1/2
3. tan ^2x + 2 tan x+1 = 0
4. cos 2x + cos x = 0
Thanks for your help so I can understand how to do these!
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! right then...crash course in trig equations. THIS IS MY METHOD - honed over years to be simple and fast and easy. What more can you ask for? ;-)
Worst thing is trying to explain it here without a diagram! Here goes.
picture a graph ie x-y axes:
Label each quadrant of the graph in your head Q1, Q2, Q3 and Q4, starting with top right being Q1, and going anticlockwise until Q4 is bottom right. Let the zero angle be the positive x-axis, and then as we travel around the graph, the +y-axis is 90 degrees, the -ve x-axis is 180 degrees, the -y axis is 270 degrees and + x-axis is 360 degrees...etc etc.
Draw 2 diagonal straight lines going through the centre, one bottom left to top right and the other bottom to top left.
the angle that the 2 straight lines make, each, in all 4 quadrants is the angle we want to find.
Now:
Q1 --> s, c, t are all +
Q2 --> s only is +
Q3 --> t only is +
Q4 --> c only is +
so, take an example...find where sinx=1/2.
Your calculator always quotes the Q1 answer..put it in...you get 30 degrees.
What about the other answer...typically there are 2 answers for every 360 degree cycle, well sin is positive too in Q2, so we need to find that angle, which is 180-30 --> 150. Put sin150 in your calculator, and you get 1/2!! Voila.
For the sake of interest, if the question had been sinx=-1/2. ALWAYS start by finding the positive version! since that is the one your calculator stores...here 30 degrees again. But since the question asks for -1/2, we need the quadrants were sin is NOT positive ie Q3 and Q4.
Q3 --> 180+30 --> 210
Q4 --> 360-30 --> 330.
Put sin210 or sin330 in your calculator and both say -1/2. DONE!
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Now for your answers:
1. 
first, find x where  on your calculator. This is x=60.
We need the negative versions of the angle for sin...Q3 and Q4.
so, x=180+60 and x=360-60
x=240, 300. Done! well...convert these to radians too :-)
2. 
First find 3x when  --> 3x = 30
We need the negatives, so again Q3 and Q4:
3x = 180+30 or 3x=360-30
3x = 210 or 3x = 330
so, x=70 or x=110
3. 
(tanx + 1)^2 = 0
so tanx + 1 = 0 --> tanx = -1
First find when tanx = +1 on your calculator... x = 45
Now find the negative versions for tan (Q2 and Q4).
180-45 = 135
360-45 = 315
Done.
4. cos2x + cosx = 0
USe the identity that 2cos^2(x) = 1 + cos2x so we get 2cos^2(x) - 1 + cosx = 0. This is another quadratic: 2cos^2(x) + cosx - 1 = 0 which factorises to (2cosx - 1)(cosx + 1) = 0
so (2cosx - 1) = 0 Or (cosx + 1) = 0
--> 2cosx = 1 or cosx = -1
we get cosx = 1/2 or -1
when is cosx = 1/2 on your calculator?..60 degree. The other Q is Q4, so other answer is 360 - 60 --> 300 degrees
When is cosx=-1...well, when is it equal to 1?...angle is zero. What about negatives? Q2, Q3....one is 180-0 and the other is 180+0...basically just one answer 180!
full answer is x=60, 180, 300.
Hope this lot helps :-)
jon.
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