solve for the exact value of x where 0 greater than or equal to X and less than/equal to 2 pi
Solve & plot a graph for the following: F(X)=cos(X)-cos to second power X for the interval negative pi greater than or equal to X less than or equal to positive pi
Solve: 2 cos2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi.
Solve: sin x - 2sin x cos x = 0 for 0 is less than or equal to x is less than 2pi.
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 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 for x in the equation 3sin^2x = cos^2x; 0
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.
For 2cos^2X + cosX - 1= 0 find X 0<X<360 <=greater than or equal to.
trig question2sin^2x+sinx-6=0how to solve for x when the restriction is x is greater than or equal to zero but less than 2pi
If cosx=sqrt{3}/2 and −pi/2<=x<=0 then the exact value of x, in radians
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