Solve: tanx + sec2x = 1

Question

with intervals of : [ -pi/2 ≤ x ≤ pi/2 ]

Answer 1

This could be tan(x)+sec²(x)=1 or tan(x)+sec(2x)=1.

(1) tan(x)+sec²(x)=1

tan²(x)=sec²(x)-1, so tan(x)+tan²(x)=0=tan(x)(1+tan(x)).

So tan(x)=0⇒x=0 or tan(x)=-1⇒x=-¼π.

(2) tan(x)+sec(2x)=1

tan(x)+1/(2cos²(x)-1)=1,

tan(x)+1/(2/sec²(x)-1)=1,

tan(x)+1/(2/(1+tan²(x))-1)=1,

tan(x)+(1+tan²(x))/(1-tan²(x))=1,

tan(x)-tan³(x)+1+tan²(x)=1-tan²(x),

tan(x)-tan³(x)+2tan²(x)=0,

-tan(x)+tan³(x)-2tan²(x)=0,

tan(x)(tan²(x)-2tan(x)-1)=0, so tan(x)=0⇒x=0, or:

tan²(x)-2tan(x)=1,

tan²(x)-2tan(x)+1=2,

(tan(x)-1)²=2,

tan(x)=1±√2⇒x=⅜π (67.5°) or -⅛π (-22.5°).