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 = 0 cos^2x - 1 - (1 - cos^2x) = 0 cos^2x - 1 - 1 + cos^2x = 0 2cos^2x - 2 = 0 (2cos^2x/2)= (-2/2) cos^
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 _____ ?
Solve each equation for 0 is less than and equal to "x" is less than and equal to 3603sinx = 2cos^{2}x-----I don't know how to solve this equation...this is what I have, but I don't know if I'm on the right track or not3sinx = 1 - 2sin^{2}x3sinx - 1 + 2sin^{2}x = 02sin^{2}x + 3sinx - 1 = 0sinx(3 + 2sinx) = 1...and if I am, what do I do next now?-----The answers are:30 and 150 degrees
For 2cos^2X + cosX - 1= 0 find X 0<X<360 <=greater than or equal to.
Solve: 2 sin x - 1 = 0 for 0 degrees is less than or equal to x is less than 360 degrees.
9(tan(2a)+2cos(a)=0 interval 0 less than or equal to a less than or equal to 2piSolve each equation...trig equations for linear sin and cos
Solve each equation on the interval:0 is less than and equal to theta is less than and equal to 360--2sin2x = 1-----My Answers:Let y represent 2x2siny = 12siny/2 = 1/2siny = 1/2y = sin^-1(1/2)y = 302x = 302x/2 = 30/2x = 15 (Quadrant 1)Quadrant 2180 - theta= 180 - 30= 1502x = 1502x/2 = 150/2x = 75Therefore, x = 15 and 75 degrees---Book's Answers:15, 75, 195, 255---My Problem:I'm asking how do you get 195 and 255 degrees if the equation...
Find the value of x in the interval 0 is less than or equal to x and 360 is greater than or equal to x which satisfies the equation cos x - 2 cos x sin x.
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 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
solve the equationsin(3x) + (1/2) = 2cos^2(x)for X to be between 0 and 360
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