Math Homework Help

• Math
First transform `sin^2(x)` into `1-cos^2(x)` and obtain `2-2cos^2(x)+5cos(x)=4,` `2cos^2(x)-5cos(x)+2=0.` This is a quadratic equation for cos(x), the roots are 2 and 1/2. cos(x) never =2....

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• Math
First transform `cos^2(x)` into `1-sin^2(x)` : `2-2sin^2(x)+7sin(x)=5,` `2sin^2(x)-7sin(x)+3=0.` This is a quadratic equation for sin(x), the roots are sin(x)=3 and sin(x)=1/2. The first is...

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• Math
`cot^2(x)=9,` cot(x)=3 or cot(x)=-3. The general solution for cot(x)=a is On `(0, 2pi)` there are `cot^(-1)(3), cot^(-1)(3)+pi, pi-cot^(-1)(3), 2pi-cot^(-1)(3).` Note that `cot^(-1)(-3)=-cot^(-1)(3).`

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• Math
This is a quadratic equation for cot(x). There are two roots, cot(x) = 1 and cot(x)=5. Now solve these equations. On `(0, 2pi)` `cot(x)=1` at `x_1=pi/4` and `x_2=(5pi)/4.` Also cot(x)=5 at...

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see attached graph. Solutions in the given interval are, x `~~` 0.500 , 0.700 , 2.400 , 2.600

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• Math
`12sin^2(x)-13sin(x)+3=0` using quadratic formula, `sin(x)=(-(-13)+-sqrt((-13)^2-4*12*3))/(2*12)` `sin(x)=(13+-sqrt(169-144))/24` `sin(x)=(13+-sqrt(25))/24` `sin(x)=(13+-5)/24=3/4,1/3` Solutions of...

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• Math
`3tan^2(x)+4tan(x)-4=0` using quadratic formula, `tan(x)=(-4+-sqrt(4^2-4*3*(-4)))/(2*3)` `tan(x)=(-4+-sqrt(16+48))/6` `tan(x)=(-4+-8)/6=-2 , 2/3` solutions for tan(x)=-2 for the range...

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• Math
`tan^2(x)+3tan(x)+1=0` using quadratic equation formula, `tan(x)=(-3+-sqrt(3^2-4*1*1))/2` `tan(x)=(-3+-sqrt(9-4))/2` `tan(x)=(-3+-sqrt(5))/2` Solutions for `tan(x)=(-3-sqrt(5))/2` for the range...

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• Math
`4cos^2(x)-4cos(x)-1=0` using quadratic formula, `cos(x)=(-(-4)+-sqrt((-4)^2-4*4*(-1)))/(2*4)` `cos(x)=(4+-sqrt(16+16))/8` `cos(x)=(1+-sqrt(2))/2` For cos(x)=`(1+sqrt(2))/2` , No solution since...

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• Math
`tan^2(x)+tan(x)-12=0` using quadratic formula, `tan(x)=(-1+-sqrt(1^2-4*1*(-12)))/2` `tan(x)=(-1+-sqrt(49))/2=(-1+-7)/2=3,-4` solutions for tan(x)=3 for the range `0<=x<=2pi` `x=arctan(3) ,...

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This is the quadratic equation for tan(x). It has roots -1 and 2. tan(x)=-1 on `(0, 2pi)` at `x_1=(3pi)/4` and `x_2=(7pi)/4.` tan(x)=2 on `(0, 2pi)` at `x_3=tan^(-1)(2)` and `x_4=pi+tan^(-1)(2).`...

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• Math
`y= sin ((pix)/2) + 1` Before we solve for the x-intercepts, let's determine the period of this function. Take note that if a trigonometric function has a form y= Asin(Bx + C) + D, its period...

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• Math
You need to evaluate the x intercepts of the graph, hence, you need to remember that the graph intercepts x axis at y = 0. Hence, you need to solve for x the equation `y= f(x) = 0` . `sin pi*x +...

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• Math
x-intercepts are points where y=0, so we have to solve the equation `tan^2((pix)/6)-3=0.` This implies that `tan((pix)/6)=sqrt(3)` or `tan((pix)/6)=-sqrt(3),` and this in turn implies...

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• Math
x-intercepts are the points where y=0. So we have to solve equation `sec^4((pix)/8)-4=0,` or `sec^4((pix)/8)=4.` Then `sec^2((pix)/8)=2,` or `sec((pix)/8)=+-sqrt(2).` sec(y) = 1/cos(y), therefore...

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• Math
Solutions of `2sin(x)+cos(x)=0` in the interval (0,2pi) See the attached graph, x `~~` 2.700 , 5.800

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• Math
See the attached graph Solutions in the given interval are, `x=pi/4 , (3pi)/4, (5pi)/4 , (7pi)/4 , (7pi)/6 , (11pi)/6` `x~~0.800 , 2.300 , 3.600 , 3.900 , 5.500 , 5.800`

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• Math
See attached graph x=`pi/3 , (5pi)/3` x `~~` 1.000 , 5.200

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• Math
See the attached graph. Since the equation is undefined for x=pi/2 x=pi/6 , 5pi/6 x `~~` 0.500 , 2.600

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• Math
see attached graph solutions in the given interval are, x `~~` 0.875 , 3.375

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see attached graph Solution in the given interval, x `~~` 4.900

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see the attached graph. Solutions in the given interval are, x `~~` 0 , 2.700 , 3.100(pi) , 5.800 , 6.200(2pi)

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• Math
See the attached graph. Solutions in the given interval are, x `~~` 0.500 , 2.700 , 3.650 , 5.850

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• Math
See attached graph Solutions in the given interval are, x `~~` 1.000 , 1.750 , 4.125 , 4.875

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• Math
Find all solutions to the equation `2cos^2(x)+cos(x)-1=0` in the interval `[0,2pi).` `2cos^2(x)+cos(x)-1=0` `(2cos(x)-1)(cos(x)+1)=0` Set each factor equal to zero and solve for the x value(s)....

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• Math
Solve the equation `2sin^2(x)+3sin(x)+1=0` in the interval [0,2pi). `2sin^2(x)+3sin(x)+1=0` `(2sin(x)+1)(sin(x)+1)=0` Set each factor equal to zero and solve for the x value(s). `2sin(x)+1=0`...

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Find all solutions to the equation `2sec^2(x)+tan^2(x)-3=0` in the interval `[0,2pi).` `2sec^2(x)+tan^2(x)-3=0` Use the pythagorean identity `sec^2(x)=1+tan^2(x)` to substitute in for...

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• Math
`cos(x)+sin(x)tan(x)=2` `cos(x)+sin(x)(sin(x)/cos(x))=2` `(cos^2(x)+sin^2(x))/cos(x)=2` `1/cos(x)=2` `cos(x)=1/2` General solutions for cos(x)=1/2 are, `x=pi/3+2pin , (5pi)/3+2pin` Solutions for...

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• Math
`csc(x)+cot(x)=1` `1/sin(x)+cos(x)/sin(x)=1` `1+cos(x)=sin(x)` `1+cos(x)-sin(x)=0` `1+cos(x)-cos(pi/2-x)=0` ` ` `1+2sin((x+pi/2-x)/2)sin((pi/2-x-x)/2)=0` `1+2sin(pi/4)sin(pi/4-x)=0`...

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• Math
`sec(x)+tan(x)=1` `1/cos(x)+sin(x)/cos(x)=1` `1+sin(x)=cos(x)` `1+sin(x)-cos(x)=0` `1+cos(pi/2-x)-cos(x)=0` `1+(2sin((pi/2-x+x)/2)sin((x-pi/2+x)/2)=0` `1+2sin(pi/4)sin(x-pi/4)=0`...

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• Math
`2cos(2x)-1=0,` `cos(2x)=1/2.` The general solution is `2x = +-arccos(1/2)+2kpi,` `kinZZ.` Because `arccos(1/2) = pi/3,` the final answer is `x = +-pi/6+kpi,` `kinZZ.`

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`2sin(2x)+sqrt(3)=0,` `sin(2x) = -sqrt(3)/2.` The general solution is `2x = (-1)^k*arcsin(-sqrt(3)/2)+kpi,` `kinZZ.` Because `arcsin(-sqrt(3)/2) = -pi/3,` the final answer is...

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• Math
tan(3x)=1. tan(y) has period `pi` and reaches any value once on each period. We know that `tan(pi/4) = 1,` so `3x = pi/4 + k*pi, kinZZ.` The answer is `x=pi/12 + k*(pi/3), kinZZ.`

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`sec(4x)-2=0,` `sec(4x)=2.` Because `sec(x)=1/cos(x),` obtain `cos(4x) = 1/2.` The general solution is `4x=+-arccos(1/2)+2kpi,` `kinZZ.` `arccos(1/2) = pi/3,` so the final answer is...

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`2cos(x/2)-sqrt(2)=0,` `2cos(x/2)=sqrt(2),` `cos(x/2) = sqrt(2)/2.` The general solution is: `x/2 = +-arccos(sqrt(2)/2) + 2kpi,` `kinZZ.` `arccos(sqrt(2)/2) = pi/4,` so the final answer is: `x =...

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`sin(x/2)=-sqrt(3)/2.` The general solution is `x/2=(-1)^k*arcsin(-sqrt(3)/2)+kpi, kinZZ.` So `x=(-1)^(k+1)*(2pi/3)+2kpi, kinZZ,` because `arcsin(-sqrt(3)/2)=-pi/3.`

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• Math
`tan(3x)tan(x-1)=0` solving each part, tan(3x)=0 General solutions for tan(3x)=0 are, `3x=0+pin` `x=(pin)/3` General solutions for tan(x-1)=0 are, `x-1=0+pin` `x=1+pin` so the solutions are,...

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• Math
`cos(2x)(2cos(x)+1)=0` solving each part, cos(2x)=0 General solutions for cos(2x)=0 are, `2x=pi/2+2pin , (3pi)/2+2pin` `x=(pi+4pin)/4 , (3pi+4pin)/4` Solving 2cos(x)+1=0, `2cos(x)=-1`...

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• Math
Solve `sin(x)(sin(x)+1)=0` `sin(x)=0` `x=0+pin` `sin(x)+1=0` `sin(x)=-1` `x=(3pi)/2+2pin`

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Solve the equation `(2sin^2(x)-1)(tan^2(x)-3)=0` Set each factor equal to zero and solve for the x values. `2sin^2(x)-1=0` `2sin^2(x)=1` `sin^2(x)=1/2` `sin(x)=+-sqrt(1/2)` `sin(x)=+-sqrt(2)/2`...

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`cos^3(x)=cos(x)` Let `cos(x)=y` `y^3=y` `y^3-y=0` `y(y+1)(y-1)=0` solve for y, y=0 , -1 , 1 Therefore cos(x)=0 , cos(x)=-1 and cos(x)=1 General solutions for cos(x)=0 are, `x=pi/2+2pin ,...

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Solve: `sec^2(x)-1=0,[0,2pi)` `tan^2(x)=1` ` ` `tan(x)=+-sqrt(1)` `tan(x)=+-1` `x=pi/4,x=(3pi)/4,x=(5pi)/4,x=(7pi)/4`

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`3tan^3(x)=tan(x)` `3tan^3(x)-tan(x)=0` `tan(x)(3tan^2(x)-1)=0 =>` `tan(x)=0 , (3tan^2(x)-1)=0` General solutions for tan(x)=0 are, `x=0+pin` Solutions in the range `0<=x<=2pi` are, `x=0...

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• Math
Find all solutions for `2sin^2(x)=2+cos(x)` in the interval [0, `2pi).` Use the pythagorean identity `sin^2(x)+cos^2(x)=1.` Solve for `sin^2(x)` and the pythagorean identity would be...

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Solve the equation `sec^2(x)-sec(x)=2,[0,2pi).` `sec^2(x)-sec(x)-2=0` `(sec(x)-2)(sec(x)+1)=0` Set each factor equal to zero and solve for the x values. `sec(x)-2=0` `sec(x)=2` `cos(x)=1/2`...

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`sec(x)csc(x)=2csc(x)` `sec(x)csc(x)-2csc(x)=0` `csc(x)(sec(x)-2)=0` `csc(x)=0 , sec(x)=2` No solutions for x for csc(x) in the range `0<=x<=2pi` General solutions for sec(x)=2 are...

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`2sin(x)+csc(x)=0` `2sin(x)+1/sin(x)=0` `(2sin^2(x)+1)/sin(x)=0` solving above , `2sin^2(x)+1=0` `2sin^2(x)=-1` `sin^(2)x=-1/2` `sin(x)=+-i/sqrt(2)` Solutions for the range `0<=x<=2pi` No...

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`sin(x)-2=cos(x)-2` `sin(x)-2-cos(x)+2=0` `sin(x)-cos(x)=0` `(sin(x)-cos(x))/cos(x)=0` `tan(x)-1=0` `tan(x)=1` General solutions for tan(x)=1 are `x=pi/4+pin` Solutions for the range...

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Solve the equation: `sqrt(3)csc(x)-2=0` `csc(x)=2/sqrt(3)` Since sin(x) and csc(x) are reciprocal functions `sin(x)=sqrt(3)/2` `x=pi/3+2pin` `x=(2pi)/3+2pin`

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