T=-1 sinA=sin(π/2-3A), A=2nπ+π/2-3A, 4A=2nπ+π/2, A=nπ/2+π/8 where n is an integer. Also, π-A=2nπ+π/2-3A, 2A=2nπ-π/2, A=nπ-π/4. The hard way: cos3A=cos(2A+A)=cos(2A)cosA-sin(2A)sinA. Let s=sinA and c=cosA, then s²+c²=1. cos3A=(2c²-1)c-2c(1-c²)=c(4c²-3). s=c(4c²-3) is the original equation. Let t=tanA=s/c, then c²=1/(1+t²). t=4c²-3=4/(1+t²)-3=(4-3-3t²)/(1+t²)=(1-3t²)/(1+t²). So t+t³=1-3t², t³+3t²+t-1=0=(t+1)(t²+2t-1). So t=-1 is a solution. t²+2t-1=0 is a solution, t²+2t+1-1-1=0=(t+1)²-2, so t=-1+√2 and t=-1-√2 are solutions. Therefore tanA=-1, -1+√2, -1-√2 are the three solutions from which: A=-π/4, π/8, -3π/8 radians and these values +2πn where n is an integer. Replacing π by 180° converts the solutions to degrees.