Solve for x: tan^2 x - square root 3=0 and 2cos^2 x - 3sinx -3 =0 Solution (a) [tan^2(x)-sqrt(3)=0 ] [tan^2(x)=sqrt(3) ] [tan(x)=pm sqrt(sqrt(3)) ] [x=tan^(-1)sqrt(sqrt(3)) ] [x~~.9210 +npi ] or [x~~2.2206+npi ] in radians, where n is an integer. ** Was this supposed to be [tan^2(x)-3=0 ] ? If so, [tan(x)=pm sqrt(3) ] and [x=pi/3+npi ] or [x=(2pi)/3+npi ] .** (b) [2cos^2(x)-3sin(x)-3=0 ] This looks almost like a quadratic in sin(x), so we rewrite the term with cosine squared in terms of sine using the pythagorean relationship. [cos^2(x)+sin^2(x)=1 ==> cos^2(x)=1-sin^2(x) ] so: [2(1-sin^2(x))-3sin(x)-3=0 ] [-2sin^2(x)-3sin(x)-1=0 ] [2sin^2(x)+3sin(x)+1=0 ] [(2sin(x)+1)(sin(x)+1)=0 ] [sin(x)=-1/2 ] or sin(x)=-1.