Trigonomerty

Help please

cos(5x+100) = ? ....... work it out,
5x+10 = ? ................. work it out,
5x = ? ....................... work it out,
x = ? ......................... work it out.

Der:

could you please solve 2 cos (5x + 10) = 1 For 0< x < 360

You can see from below that there are several solutions that lie within the range 0<x<360

angles.PNG

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

My calculation above assumed that 5x+10 was equal to theta measured in radians. However, it could also be that it should be equal to theta measured in degrees, in which case the results would look different (but there would still be several values of x in the range 0 to 360).

Der wrote:could you please solve 2 cos (5x + 10) = 1 For 0< x < 360

cos(5x+10) = 0.5
cos(5x+10) - 0.5 = 0
If I graph y = cos(5x+10) - 0.5
And y=0 (the x axis)
Where they intersect will be the answers
And, they are the same as the ones I got by hand. Now I am happy

140302 y=cos(5x+10)-0.5.JPG

With 5x+10 equal to angle in degrees then with cos(angle)=1/2 we must have 5x+10 in the form 360n+60 or 360n-60, where n is an integer. That is, x must be in the form 72n+10 or 72n-14. With n = 0 only the first form gives x in the range 0 to 360; with n = 5 only the second form gives x in the right range; with n = 1, 2, 3, and 4, both forms give x in the right range. The values I get for x this way are: 10, 58, 82, 130, 154, 202, 226, 274, 298, 346, which match those in Melody's picture. Now I'm happy too, Melody!