Question

Medal always given, help is always appreciated(: How would I solve -sin^2x=2cosx-2 ??

Answer 1

Hey Velasquez.....do u knw
cosx=1-2sin^2x/2

Answer 2

like a first step
sin^2 x +cos^2 x =1 so
sin^2 x = 1-cos^2 x and now rewrite sin^2 x =2cosx -2

1 -cos^2 x =2cos x -2

cos^2 x +2cos x -3 =0 than note cos x = t and rewrite it will give you

t^2 +2t -3 = 0 so now this you need to solve it for t and wll get like roots t_1 and t_2
what dont forget you need calculing x making cos x = t_1 and cos x =t_2

hope this will help you

Answer 3

ladkiyo ko sirf qn puchna aata hai

Answer 4

stuck on the last part :/

Answer 5

quadratic formula

Answer 6

oh the last part

Answer 7

\(\bf sin^2(x)=2cos(x)-2\\ \quad \\
\textit{recall that }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\qquad thus\\ \quad \\
sin^2(x)=2cos(x)-2\implies 1-cos^2(x)=2cos(x)-2\\ \quad \\
0=cos^2(x)+2cos(x)-1\)

factor the quadratic like you'd any other, to get the roots and thus the values for cos(x)

Answer 8

thanks

Answer 9

@jdoe0001 how you have got this 0=cos2(x)+2cos(x)−1 ?

so because 1 on the right side will be minus 1 and with -2 will add -3

yes ?