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the problem is 2cos^2x +

the problem is2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
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2. asked by alex
3. 859 views
cosA= 5/9 find cos1/2Aare you familiar with the half-angle formulas? the one I would use here is cos A = 2cos^2 (1/2)A - 1 5/9 +
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2. asked by Anonymous
3. 810 views
What am I doing wrong?Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) =
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2. asked by Anonymous
3. 966 views
Determine the solution of the equetion: 2cos x-1=0? 2cos x=0? and 2cos x+1=0?
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2. asked by Miley
3. 493 views
how does 2cos(2x)-2cos(x)sin(x) becomes 2cos(2x)-sin(2x)? Try to explain each step simply.
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2. 70 views
Identify the equation for the following graph: (1 point) Responses y=2cos(2x) y is equal to 2 cosine 2 x y=−2cos(2x) y is
1. 1 answer
2. 91 views
Multiply then use fundamental identities to simplify the expression below and determine which of the following is not equivalent
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2. asked by Tristian
3. 1,672 views
Which cosine function has maximum of 2, a minimum of -2, and a period of 2pi/3?Choose from answers below, correct one please. y
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2. 46 views
Please help find the solution to the initial value problem dy/dx=(1+2cos^2 x)/y with(y>0), and y=1 when x=0. Thanks.
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2. asked by John
3. 444 views
give the solution of the initial-value problemdx/dy=(1+2cos^2(x))/y, (y > 0), y= 1 when x = 0. Thanks.
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2. asked by olek
3. 509 views