Q: Given 2sin s equals cos s find all possible values of sin s?

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sin4x=(4sinxcosx)(1-2sin^2x)

3 sin(x) = 2sin(x) = 2/3x = 41.81 degreesx = 138.19 degrees

2sin(x+1)=3cos(x-1)Since there is only one variable in this equation, it cannot be implicitly differentiated.

2sin(x), or 5.4cos(y), tan(z)

R^2sin(theta)d(theta)d(phi)(r-hat)

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sin4x=(4sinxcosx)(1-2sin^2x)

2 pi

3

3 sin(x) = 2sin(x) = 2/3x = 41.81 degreesx = 138.19 degrees

2sin(y) = 2x/sqrt(1+x^2)

Placing a question mark at the end of a phrase does not make it a sensible question. Try to use a whole sentence to describe what it is that you want answered.

sin 2x + cos x = 0 (substitute 2sin x cos x for sin 2x)2sin x cos x + cos x = 0 (divide by cos x each term to both sides)2sin x + 1 = 0 (subtract 1 to both sides)2sin x = -1 (divide by 2 to both sides)sin x = -1/2Because the period of the sine function is 360⁰, first find all solutions in [0, 360⁰].Because sin 30⁰ = 1/2 , the solutions of sin x = -1/2 in [0, 360] arex = 180⁰ + 30⁰ = 210⁰ (the sine is negative in the third quadrant)x = 360⁰ - 30⁰ = 330⁰ (the sine is negative in the fourth quadrant)Thus, the solutions of the equation are given byx = 210⁰ + 360⁰n and x = 330⁰ + 360⁰n, where n is any integer.

y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)

2sin(x+1)=3cos(x-1)Since there is only one variable in this equation, it cannot be implicitly differentiated.

10

(/) = theta sin 2(/) = 2sin(/)cos(/)

The answer will range between '2' & '-2' Reason; The Sine function ranges between '1' & '-1' , so if it has a coefficient of '2', this will increase the range to '2' & '-2'.