Der: could you please solve 2 cos (5x + 10) = 1 For 0< x < 360
cos(5x+10) = 1/2
Firstly cos is positive in the 1st and the 4th quadrants (I use "ALL Stations To Central" to remember this)
I have drawn a equilateral triangle of side length 2 units. (Angles are all 60 degrees)
I have then bisected it to form 2 congruent 30,60 90 degree triangles
The height is sqrt(3)
I draw this every time I want trig ratios for 30 and 60 degree angles.
I recognise cos(theta)=1/2 as one such ratio.
From the diagram it can be seen that cos(60) = 1/2
First quadrant is 60 degrees+360n, 4th is 360-60=300+360n degrees (Where an is in the set of integers)
5x+10=60, 420, 780, 1140, 1500
5x = 50, 410, 770, 1130, 1490
x = 10, 82, 154, 226, 298,
5x+10 = 300, 660, 1020,1380, 1740
5x = 290, 650, 1010, 1370, 1730
x = 58,130, 202, 274, 346
So the answers (in degrees) are x = 10,58, 82, 130, 154, 202, 226, 274, 298, 346
My answer and Alan's answer still don't mesh. I'll do some more homework. This is fun.
140302 30,60,90 triangle.JPG