Question

Sin^6x+cos^6x+ksin^22x=1 than k=?

Answer 1

Given: $sin^{6}x+cos^{6}x+k\:sin^{2}2x=1$

To find: the value of $k$

Solution:

Now, $sin^{6}x+cos^{6}x+k\:sin^{2}2x=1$

$\Rightarrow (sin^{2}x)^{3}+(cos^{2}x)^{3}+k\:sin^{2}2x=1$

$\Rightarrow (sin^{2}x+cos^{2}x)(sin^{4}x+cos^{4}x-sin^{2}x\:cos^{2}x)+k\:sin^{2}2x=1$

• using the identity:
• $a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)$

$\Rightarrow (sin^{2}x)^{2}+(cos^{2}x)^{2}-sin^{2}x\:cos^{2}x+k\:sin^{2}2x=1$

$\Rightarrow (sin^{2}x+cos^{2}x)^{2}-2\:sin^{2}x\:cos^{2}x-sin^{2}x\:cos^{2}x+k\:sin^{2}2x=1$

• since $sin^{2}x+cos^{2}x=1$
• and $a^{2}+b^{2}=(a+b)^{2}-2ab$

$\Rightarrow 1-3\:sin^{2}x\:cos^{2}x+k\:sin^{2}2x=1$

$\Rightarrow -3\:sin^{2}x\:cos^{2}x+k\:(2\:sinx\:cosx)^{2}=1$

• since $sin2x=2\:sinx\:cosx$

$\Rightarrow -3\:sin^{2}x\:cos^{2}x+4k^{2}\:sin^{2}x\:cos^{2}x=0$

• since $(a.b.c....)^{p}=a^{p}.b^{p}.c^{p}....$

$\Rightarrow -3+4k=0$

• since $sinx\:cosx\neq 0$

$\Rightarrow 4k=3$

$\Rightarrow k=\frac{3}{4}$

Answer: the value of $k$ is $\frac{3}{4}$