CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The solution of $$\displaystyle \sin ^{ 8 }{ x } +\cos ^{ 8 }{ x } =\frac { 17 }{ 32 } $$ is


A
nπ2±π8
loader
B
nπ±π4
loader
C
nπ±π8
loader
D
No solution
loader

Solution

The correct option is A $$\displaystyle \frac { n\pi  }{ 2 } \pm \frac { \pi  }{ 8 } $$

$$\sin ^{8}(x)+\cos ^{8}(x)$$

$$=(\sin ^{4}x+\cos ^{4}x)^{2}-2\sin ^{4}x\cos ^{4}x$$

$$=((\sin ^{2}x+\cos ^{2}x)^{2}-2\sin ^{2}x\cos ^{2}x)^{2}-2\sin ^{4}x\cos ^{4}x$$

$$=(1-\dfrac{\sin ^{2}2x}{2})^{2}-\dfrac{\sin ^{4}2x}{8}$$

 

$$=1+\dfrac{\sin ^{4}2x}{4}-\sin ^{2}2x-\dfrac{\sin ^{4}2x}{8}$$

 

$$=1+\dfrac{\sin ^{4}2x}{8}-\sin ^{2}(2x)$$
$$=\dfrac{17}{32}$$

Therefore 

Let $$\sin ^{2}(2x)=t$$
Hence

$$\dfrac{t^{2}}{8}-t+\dfrac{15}{32}=0$$

 

$$4t^{2}-32t+15=0$$

$$\Rightarrow t=\dfrac{32\pm\sqrt{1024-240}}{8}$$

 

$$\Rightarrow t=\dfrac{32\pm4\sqrt{64-15}}{8}$$

 

$$\Rightarrow t=\dfrac{8\pm\sqrt{49}}{2}$$

 

$$\Rightarrow t=\dfrac{15}{2}$$ or $$t=\dfrac{1}{2}$$

 

Now, $$t=\dfrac{15}{2}$$ is not possible.

 

$$\sin ^{2}(2x)\epsilon[0,1]$$

Therefore 

$$\sin ^{2}2x=\dfrac{1}{2}$$

$$\sin (2x)=\pm\dfrac{1}{\sqrt{2}}$$

 

$$2x=\dfrac{(2n+1)\pi}{4}$$

Or 

$$2x=\dfrac{n\pi}{2}\pm\dfrac{\pi}{8}$$

$$\Rightarrow x=\dfrac{n\pi}{4}\pm\dfrac{\pi}{8}$$


Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image