IIT JEE , AIEEE Forum - Mathematics , Trignometry

hsbhatt

Solve the equation for x[0, 2):

$\sin^6 x + \cos^6 x = \frac{1}{4}$

mukulaish

sin^6x+cos^6x=1/4

(sin^2x+cos^2x){1-3sin^2xcos^2x}=1/4

1-3sin^2xcos^2x=1/4
1/4=sin^2xcos^2x
sinxcosx=1/2 or sinxcos=-1/2
sin2x=1 or sin2x=-1
2x=pi/2 or 2x=-pi/2
x=pi/4 or x=-pi/4

m i correct???

sriram.a

(1-3(sinxcosx)^2)=1/4

(sinxcosx)^2=1/4

(sin2x)^2=1

sin2x=+-1

hence x=(pi/4,3pi/4,5pi/4,7pi/4)

if iam right plzzz rate me cheeeeerrrsss

shinee

where nbelongs to N

and second equation is obtained by replacing pi/2 with 3pi/2

hsbhatt

An alternative way of doing this problem:

$\text{The Power Mean Inequality states that:} \\ \\ \left ( \frac{a^3+b^3}{2} \right )^\frac{1}{3} \ge \frac{a+b}{2} \\ \\ \text{Here, use} \ a = \sin^2 x; b= \cos^2 x \\ \\ \text{We get} \ \sin^6 x + \cos^6 x \ge \frac{1}{4} \\ \\ \text{Here, equality holds} \ \Rightarrow a = b \\ \\ \text{i.e.} \ \sin^2 x = \cos^2 x \Rightarrow \tan x = \pm 1 \\ \\ \text{This gives 4 solutions for} \ x \in [0, 2 \pi]: \\ \\ \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{ 5 \pi}{4} \ \text{and} \ \frac{7 \pi}{4} $

ravi00

itz simply (2n-1)/4.

since sin⁶x+cos⁶x =1/4

(sin²x)³+(cos²x)³=1/4

by solving we get

1/8+1/8=1/4

hsbhatt

i think you meant substitute not solve.

little_genius

so sin^6x+cos^6x

=(s^x+c^x)(s^4x-s^2xc^2x+c^4x)

=(1-2sin^2x.cos^2x)

=1-(1/2)(sin4x)

now

1-(1/2)(sin4x) =1/4

so , (1/2)(sin4x)=3/4

so,sin4x=3/2

which is impossible...

so no value of x satisfies the eqn!!

little_genius

where i m wrong....plzz point it out..

hsbhatt

see sriram's solution above

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