solving these mindboggling trig stuff

The second isn't too bad..we have

 (b)  sinpcosp= 1/2

Note that sinpcosp is really just (1/2)(2)sinpcosp   = (1/2)sin(2p)

So we have

(1/2)sin(2p) = 1/2        multiply through by 2

sin(2p) = 1

Since sinp = 1 at  90 and 450, then sin(2p) = 1 at 45 and 225

Here's the graph of this one......https://www.desmos.com/calculator/f1wb8idmiy

For a, we have

sin(p+15degrees)= 3cos(p+15degrees)   ...note by an identity sinA = cos(90-A)

So......let A = p + 15   and we have

sin(p+ 15) = cos(90- (p+ 15)) = cos (75 - p)....so

cos(75 - p)  = 3cos(15 + p)...and we have

cos75cosp + sin75sinp = 3[cos15cosp - sin15 sinp]  rearranging, we have

3cos15cosp - cos75cosp =  sin75sinp + 3sin15sinp

cosp(3cos15-cos75) = sinp(sin75 + 3sin15)    rearrange again

sinp / cos p = (3cos15 - cos75)/(sin75 + 3 sin15)

tanp = (3cos15 - cos75)/(sin75 + 3 sin15)

tan^-1 (3cos15 - cos75)/(sin75 + 3 sin15) = p = 56.56505117705 degrees...this could also be a 3rd quad angle = (180 + 56.56505117705) degrees = 236.56505117705 degrees

This one was definitely tougher !!

Here's the graph .....  https://www.desmos.com/calculator/8qdesoejgk

Slightly simpler way of doing (a) is to divide both sides by cos(p+15°) to get tan(p+15°) = 3

p+15° = tan^-1(3)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)} = {\mathtt{71.565\: \!051\: \!177\: \!078^{\circ}}}$$

so p = 71.565° - 15° 

or p = 56.6° (to one decimal place)

That's the first quadrant solution.  Add 180° to get the third quadrant solution.

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That ain't "slightly simpler".....that's WAY more simple.....thanks, Alan!!!.....(Melody's "Method of Over-Complication" is rubbing off on me....!!!!)

Yes that proves it - I'm a great teacher!!     LOL