Solve for x where x is a real number, sin 2x = sin x, 0 less than or equal to x less than or equal to 2 pi
cos^2x - sin^2x = sin x with x greater than negative pi and less than or = pi
solve for the exact value of x where 0 greater than or equal to X and less than/equal to 2 pi. cos^2 x-1=0 I did the factoring and FOILing and The two answers I got were: cosx=-1 cosx=1 x=pi x=0,2pi According to my instructor cosx=1's answers are not completly correct. I dont see what the problem is since cos(0) and cos(2pi) both =1. Am I missing something here? Thank you for your help.
Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360--cos^2x - 1 = sin^2x--Attempt:cos^2x - 1 - sin^2x = 0cos^2x - 1 - (1 - cos^2x) = 0cos^2x - 1 - 1 + cos^2x = 02cos^2x - 2 = 0(2cos^2x/2)= (-2/2)cos^2x = -1cosx = square root of -1And I can't do anything with this now...what am I doing wrong?--Textbook answers:0, 180, 360--
Solve: 2 sin x - 1 = 0 for 0 degrees is less than or equal to x is less than 360 degrees.
Solve: sin x - 2sin x cos x = 0 for 0 is less than or equal to x is less than 2pi.
Solve the equation if 0 is less than or equal to x and x is less than 2 pie 2sin^2x+3 cos x -3 0
Solve each equation for o is less than and/or equal to theta is less than and/or equal to 360--sin^2x = 1 = cos^2x--Work:cos^2x - cos^2x = 00 = 0--Textbook Answers:90 and 270--Btw, how would you isolate for cos^2x = 0? Would it be...x = cos^-1 square root of _____ ?
Integral from 0 to pi/6 of (sin x + cos x) less than or equal to pi(square root 3 + 1)/12
Solve for x in the equation 3sin^2x = cos^2x; 0
Solve the question: sin(2x)=cos(x), for values of x between 0 and 2 Pi