#1

2sin^2x = cosx + 1

i need to solve for X. the answers are 60, 180, and 360 but i need to know how because i have to test over this tomorrow and I can't do it =[

i need to solve for X. the answers are 60, 180, and 360 but i need to know how because i have to test over this tomorrow and I can't do it =[

#2

theres such thig as algebra 4? Our school only has up to algebra 2. unless your in college or something.

#3

Study and memorize the unit circle. It's odd your doing Trig in an Algebra class, but at the same time its odd there is an Algebra IV at your school.

#4

omg wtf i though it stoped at alg 2. damn i hate math

#5

theres such thig as algebra 4? Our school only has up to algebra 2. unless your in college or something.

in our school, it goes..

Algebra 1, Plane Geometry, Algebra 2/Trigonometry, Algebra 4/Precalc, Calculus.

I don't get it either.

#6

Study and memorize the unit circle.

i know the unit circle but how does that help me in THIS problem?

#7

We just call it Pre-Calc in my school.

#8

in our school, it goes..

Algebra 1, Plane Geometry, Algebra 2/Trigonometry, Algebra 4/Precalc, Calculus.

I don't get it either.

so algebra 4 is just another name for precal? Well we have precal so i guess its the same.

#9

so algebra 4 is just another name for precal? Well we have precal so i guess its the same.

i guess.

to the subject, can anyone help?

#10

Uh, what happened to algebra 3?

Anyway, I have no clue (specifically). This probably won't help you, but all I know about how they relate is if a sine and cosine of two numbers that add up to 90 are multiplied, it equals 1. Maybe you can manipulate that somehow.

Hold on, is that an exponent after the sine?

Anyway, I have no clue (specifically). This probably won't help you, but all I know about how they relate is if a sine and cosine of two numbers that add up to 90 are multiplied, it equals 1. Maybe you can manipulate that somehow.

Hold on, is that an exponent after the sine?

#11

Obviously your school needs to study their math because they forgot #3.

EDIT: Damn, beat.

EDIT: Damn, beat.

#12

2sin^2x = cosx + 1

Ok, you use the unit circle to find what different values equal when you take sin/cos/tan of them etc. Also look out for identies (cosx + 1.)

Ok, you use the unit circle to find what different values equal when you take sin/cos/tan of them etc. Also look out for identies (cosx + 1.)

#13

sin x / cos x = tan x

cos x / sin x = cot x = 1 / tan x

sec x = 1 / cos x

csc x = 1 / sin x

sin2 x + cos2 x = 1

tan2 x + 1 = sec2 x = 1 / cos2 x

cot2 x + 1 = csc2 x = 1 / sin2 x

I'm too lazy to do the work for you, so use these relationships until you can get a single sin/cos/tan/sec/csc/cot on one side and a number on the other side

cos x / sin x = cot x = 1 / tan x

sec x = 1 / cos x

csc x = 1 / sin x

sin2 x + cos2 x = 1

tan2 x + 1 = sec2 x = 1 / cos2 x

cot2 x + 1 = csc2 x = 1 / sin2 x

I'm too lazy to do the work for you, so use these relationships until you can get a single sin/cos/tan/sec/csc/cot on one side and a number on the other side

#14

2sin^2x = cosx + 1

Ok, you use the unit circle to find what different values equal when you take sin/cos/tan of them etc. Also look out for identies (cosx + 1.)

cosx+1 = -sinx, right?

thanks

#15

2sin^2x=cosx+1

using the identity 1=cos^2x+sin^2x

2(1-cos^2x)=cosx+1

2-2cos^2x=cosx+1

2cos^2x+cosx-1=0

if you make y=cosx we have

2y^2+y-1=0

and you solve the polynomial , then you find the values of x with

x=arccos(y)

I'm not completely sure though, make sure you study the trig identities they help a lot

using the identity 1=cos^2x+sin^2x

2(1-cos^2x)=cosx+1

2-2cos^2x=cosx+1

2cos^2x+cosx-1=0

if you make y=cosx we have

2y^2+y-1=0

and you solve the polynomial , then you find the values of x with

x=arccos(y)

I'm not completely sure though, make sure you study the trig identities they help a lot

#16

2sin^2x=cosx+1

using the identity 1=cos^2x+sin^2x

2(1-cos^2x)=cosx+1

2-2cos^2x=cosx+1

2cos^2x+cosx-1=0

if you make y=cosx we have

2y^2+y-1=0

and you solve the polynomial , then you find the values of x with

x=arccos(y)

I'm not completely sure though, make sure you study the trig identities they help a lot

the answer is supposed to be in degrees

#17

I didn't know algebra went above 2

#18

you really need to study

ok so the polynomial was

2y^2+y-1=0

the solutions are y=-1 and y=.5

so since x=arcos(y)

x=180º for y=-1

x=60º for y=.5

ok so the polynomial was

2y^2+y-1=0

the solutions are y=-1 and y=.5

so since x=arcos(y)

x=180º for y=-1

x=60º for y=.5

#19

thanks dude.. its not that i didn't study its that i wasn't in class the day it was taught and its not a chapter out of the book

#20

Substitute Sin^2x and it's a cakewalk from there.