0
2 sin(x)2 - sin(x) - 1 = 0
Let Y=sin(x) then the equation is 2*Y2 - Y - 1 =0
Delta = (-1 * -1) - 4 * 2 * -1 = 9
Y = (1 + sqrt(9)) / 4 or Y = (1 - sqrt(9)) / 4
Y = 1 or Y = -1/2
Then x = Arcsin(Y) and (in radians) x = Arcsin(1) = Pi/2 +2*k*Pi or x=Arcsin(-1/2) = -Pi/6 + 2*k*Pi
where k is an integer
What else can I help you with?
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
How do you express the term 1-csc squared x all over sin x cot x to cos x?
(1 - csc2x)/(sinx*cotx) = -cot2x/sinxcotx = -cotx/sinx = -(cosx/sinx)/sinx = -cosx/sin2x = -cosx/(1-cos2x) = cosx/(cos2x - 1)
How do you solve 6sin x 1 plus 9sin x algebraically over the domain 0 x 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
Why is 2 cosx sinx 2 sinx?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc. Also use ^ to indicate powers (eg x-squared = x^2).
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
How do you solve sin x - 2 sin x is equal to 0?
sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...
What is the derivative of sin x minus cos x?
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
Integral of sec squared radical x?
integral of radical sinx
What is sine squared?
Answer 1 Put simply, sine squared is sinX x sinX. However, sine is a function, so the real question must be 'what is sinx squared' or 'what is sin squared x': 'Sin(x) squared' would be sin(x^2), i.e. the 'x' is squared before performing the function sin. 'Sin squared x' would be sin^2(x) i.e. sin squared times sin squared: sin(x) x sin(x). This can also be written as (sinx)^2 but means exactly the same. Answer 2 Sine squared is sin^2(x). If the power was placed like this sin(x)^2, then the X is what is being squared. If it's sin^2(x) it's telling you they want sin(x) times sin(x).
What is the integral of cot squared x?
ln(sinx) + 1/3ln(sin3x) + C
When does cosx times sinx times sinx equal 1?
at the angles 0 and 360 degrees, or 0 and 2pi
How do you express the term 1-csc squared x all over sin x cot x to cos x?
(1 - csc2x)/(sinx*cotx) = -cot2x/sinxcotx = -cotx/sinx = -(cosx/sinx)/sinx = -cosx/sin2x = -cosx/(1-cos2x) = cosx/(cos2x - 1)
How do you solve 6sin x 1 plus 9sin x algebraically over the domain 0 x 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
Why is 2 cosx sinx 2 sinx?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc. Also use ^ to indicate powers (eg x-squared = x^2).
How do you solve csc x-sin x equals cos x cot x?
cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.
