Daniell Phillips

2021-12-30

If $\mathrm{cos}x\cdot \mathrm{cos}2x=\frac{14}{,}\text{}x\in [0,{90}^{\circ})$ , then what is the solution of the equation?

I attempted to solve this question as follows:

$\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$

$\Rightarrow \mathrm{cos}x({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x)=\frac{1}{4}$

${\mathrm{cos}}^{3}x-\mathrm{sin}x\left(\mathrm{cos}x\mathrm{sin}x\right)$

And I got stuck here, I did not know how to continue.

I attempted to solve this question as follows:

And I got stuck here, I did not know how to continue.

xandir307dc

Beginner2021-12-31Added 35 answers

Since $x\ne 0$ , multiplying both sides by $4\mathrm{sin}x$ and using double angle formula, $\mathrm{sin}2\theta =2\mathrm{sin}\theta \mathrm{cos}\theta$ twice, it is obtained

$\mathrm{sin}4x=\mathrm{sin}x$

whence

$4x+x=\pi \Rightarrow x=\frac{\pi}{5}={36}^{\circ}$

whence

Annie Gonzalez

Beginner2022-01-01Added 41 answers

Observe that

Vasquez

Expert2022-01-08Added 669 answers

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