Question

Solve the trigonometric equations for and match them with their solutions

Answer 1

For 30 is it cos2x=1/2

Answer 2

@Pbonnette
Did you start to solve them?

Answer 3

Not real sure what to do

Answer 4

Can you pick one and start to solve for x?

Answer 5

the first one

Answer 6

Yes,,,,...,,,,

Answer 7

inverse of sin?

Answer 8

Can you share your works/steps, please?

Answer 9

@sunnnystrong

Answer 10

is it inverse of sin?

Answer 11

Okay so:
\[4\sin^2x+1=4\]
Solve for x-->
\[\sin^{-1} \sqrt{\frac{ 3 }{ 4 }}=x\]
Solve for x -->
\[\cos2x=\frac{ 1 }{ 2 }\]
\[\cos^{-1} (\frac{ 1 }{ 2 })\div2=x\]

Answer 12

@Pbonnette ... what do you think for the next 2?

Answer 13

not real sure about the sec one

Answer 14

recall that:
\[\tan(x)=\frac{ \sin(x) }{ \cos(x) }\]
\[sec(x)=\frac{ 1 }{ \cos(x) }\]
step 1: rewrite in all cos and sin

\[\frac{ 1 }{ \cos(x) }^2+\frac{ \sin(x) }{ \cos(x) }^2=7\]
expand & multiply by LCD
\[\frac{ 1 }{ \cos^2(x) }+\frac{ \sin^2(x) }{ \cos^2(x) }=7\]
\[1+\sin^2x=7\cos^2x\]
*Pythagorean Identity
\[sin^2x+cos^2x=7\cos^2x-sin^2x\]
\[0=-6\cos^2x+2sin^2x\]
*Take square root of both sides --> Find X
\[6cosx=2sinx\]
\[3cosx=sinx\]
\[3=tanx\]
\[\tan^{-1} (3)=71 deg\]

Soo.... not sure if I did this right but** It's about 60

Answer 15

Last one is easier:
\[5\sin(x)=3-\sin(x)\]
\[6\sin(x)=3\]
\[\sin^{-1} (\frac{ 1 }{ 2 })=x\]

Answer 16

@Pbonnette .... Also when you graph:

\[y1=\sec^2(x)+\tan^2(x)\]
\[y2=7\]
They intersect @ 60 degrees... So i think i just made a computational error somewhere haha

Answer 17

\[\large\rm \color{orangered}{\sec^2x}+\tan^2x=7\]Hmm I think the third one might be a tad easier if you use one of your Pythagorean Identities right away :)\[\large\rm \color{orangered}{1+\tan^2x}+\tan^2x=7\]And then combine like-terms and such.

Answer 18

@zepdrix
Thank you!! you're awesome haha

Answer 19

:p