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Mathematics 13 Online
OpenStudy (anonymous):

Find all solutions to the equation. 7 sin^2x - 14 sin x + 2 = -5

OpenStudy (anonymous):

pi/2?

OpenStudy (anonymous):

ok how do i solve this question?

OpenStudy (campbell_st):

add 7 to both sides of the equation so you have \[7\sin^2(x) - 14\sin(x) + 7 = 0\] divide each term by 7 \[\sin^2(x) -2\sin(x) + 1 = 0\] this is a quadratic equation the factorises to a perfect square hope this helps

OpenStudy (anonymous):

ok thank you guys

OpenStudy (campbell_st):

oops should have been add 5

OpenStudy (anonymous):

Wait you're both giving me two different equations i'm confused.

OpenStudy (campbell_st):

the correct equation occurs by adding 5 to both sides of the equation \[7\sin^2(x) - 14\sin(x) + 7 = 0\]

OpenStudy (anonymous):

No confusion now. I getting out while the getting is good. Sorry about that.

OpenStudy (anonymous):

ok so once i have 7sin2(x)−14sin(x)+7=0 can you help me factor it please because the sin throws me off a bit

OpenStudy (campbell_st):

ok... divide each term by 7

OpenStudy (campbell_st):

to make it easy let u = sin(x) so you have \[u^2 - 2u + 1 = 0\] can you factor this... its a perfect square

OpenStudy (anonymous):

(u-1)(u-1)?

OpenStudy (campbell_st):

yep... or (u -1)^2 = 0 reverse the substitution (sin(x) -1)^2 = 0 so you need to solve sin(x) -1 = 0 hope that makes sense

OpenStudy (anonymous):

ok i understand the factoring now. so i would get sinx=1?

OpenStudy (campbell_st):

you will... to you need to find \[x = \sin^{-1}(1)\]

OpenStudy (anonymous):

Wait im confused how did you get x=sin^-1(1)

OpenStudy (campbell_st):

its the arcsin.... or what angle gives sin a value of 1. its 90 degrees or pi/2

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

thank you for your help:)

OpenStudy (campbell_st):

hope it makes sense

OpenStudy (anonymous):

its a bit easier now thank you

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