Question

Help !!

solve the following equations for x over the interval [0, 2pi]

Answer 1

wat

Answer 2

\[\sin2x + \sqrt{2}sinx=0 \]
\[2 \cos ^{2}x+1 = 3cosx\]

Answer 3

cos2x+2=sinx
\[2\sin ^{2}x-sinx=1\]

If someone could help out that would be great :) Im a little confused how to start off

Answer 4

so looking at th 1st equation youcan use substitution for sin(2x)

a sin(2x) = sin(x)cos(x)

so the equation becomes

\[2\sin(x)\cos(x)+ \sqrt{2}\sin(x) = 0\]

so te equation can now be factored then solved.

Answer 5

rewrite the 2nd equation as

\[2\cos^2(x) - 3\cos(x) +1 = 0\]

you now have a quadratic equation that can be solvd by factoring

Answer 6

\[2\sin^2(x) - \sin(x) - 1 = 0\]

another quadratc equation that can be solved by factoring..

Answer 7

thank you that helps a lot :) for the first one would i have to to square the whole equation to get rid of the square root right?

Answer 8

cos(2x) + 2 = sin(x)

use substitution

\[\cos(2x) \cos^2(x) - \sin^2(x)\]

then writing evething in ters of sin(x)

\[\cos(2x) = 1- 2\sin^2(x)\]

so the equation can be written as

\[2\sin^2x + \sin(x) - 3 =0\]

whicis agin a quadratic quation tha can be solved by factoring

Answer 9

for the 1st equation remove the common factor of sin(x)

\[\sin(x)(2\cos(x) - \sqrt{2}) = 0\]

Answer 10

ah ok thank you very much :)

Answer 11

marmar 10 want a medal?

Answer 12

times up you get a medal