Question

how do i find the values of x when i have cos^2x = 2sinx?

Answer 1

hmmm do you know the trig identity of \(\bf sin^2(\theta)+cos^2(\theta)=?\)

Answer 2

it = 1

Answer 3

so... if we solve that equation for say \(\bf cos^2(\theta)\) what would that give us?

Answer 4

1-sin^2x

Answer 5

ohhh .. ok . so \(\bf sin^2(\theta)+cos^2(\theta)=1\implies {\color{blue}{ cos^2(\theta)}}=1-sin^2(\theta)\qquad thus\\ \quad \\
{\color{blue}{ cos^2(x)}} = 2sin(x)\implies {\color{blue}{ 1-sin^2(x)}}=2sin(x)\\ \quad \\

0=sin^2(x)+2sin(x)-1\)

and as you can see, now you're just left with a quadratic, so just solve for \(\bf sin(x)\)

Answer 6

how would i solve the problem?

Answer 7

how do i get rid of the sinx on both sides of the equation?

Answer 8

its a quadratic equation now do you see it?

Answer 9

well, think about it just like you'd be solving say \(\bf x^2+2x-1\)

Answer 10

i tried doing the 2 pairs of parantheses but that does not work

Answer 11

hmm then use the quadratic formula

Answer 12

\(\bf 0=sin^2(x)+2sin(x)-1\\ \quad \\
sin(x)= \cfrac{ - 2 \pm \sqrt { 2^2 -4(1)(-1)}}{2(1)}\)

Answer 13

\(\bf 0=sin^2(x)+2sin(x)-1\\ \quad \\
sin(x)= \cfrac{ - 2 \pm \sqrt { 2^2 -4(1)(-1)}}{2(1)}\\ \quad \\

sin(x)= \cfrac{ - 2 \pm \sqrt { 4+4}}{2(1)}\implies sin(x)= \cfrac{ - 2 \pm \sqrt { 8}}{2(1)}\\ \quad \\

sin(x)= \cfrac{ - \cancel{2} \pm \cancel{2}\sqrt { 2}}{\cancel{2}}\)

Answer 14

yes i got the same answer

Answer 15

\(\bf sin(x)= \cfrac{ - 2 \pm 2\sqrt { 2}}{2(1)}\quad \textit{taking }sin^{-1}\\ \quad \\

sin^{-1}[sin(x)]= sin^{-1}\left[-1\pm\sqrt{2}\right]\implies x=sin^{-1}\left[-1\pm\sqrt{2}\right]\)

Answer 16

\(\bf sin(x)= -1\pm\sqrt{2}\\ \quad \\

sin^{-1}[sin(x)]= sin^{-1}\left[-1\pm\sqrt{2}\right]\implies x=sin^{-1}\left[-1\pm\sqrt{2}\right]\) for that matter... anyhow, that'd give the values for "x"

Answer 17

i got Error on my calculator for one of the 2 x values

Answer 18

the other x value was about 24. 47

Answer 19

right, the negative root, because the -1 will add to it, so that'd be an extraneous value, so you'd discard that one

it gives -2.41 which is a value outside the sine function range, -1 < sine < 1

so the valid one is the positive root one

Answer 20

well \(\bf -1 \le sin(x) \le +1\)

Answer 21

he other x value was about 24. 47 \(\large \checkmark\)