Pakisagutan. Trigonometry Equations. Class 11. 1. tan 4x - 1 = 02. sec² x

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October 2022 1 14 Report

Pakisagutan. Trigonometry Equations. Class 11.

1. tan 4x - 1 = 0
2. sec² x - 1 = 0
3. cos 2x + 3 = 5 cos x

With complete solution. Kahit super haba basta tama at kumpleto ang solution. Pwedeng pics or type niyo na lang. Salamat!

Answers & Comments

devoured

Trigonometry Equations. Class 11.

1. tan 4x - 1 = 0

2. sec² x - 1 = 0

3. cos 2x + 3 = 5 cos x

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Answer:

x = π/16
x = 0, π
x = π/3

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Solving for 1. $\tan 4x-1=0$

By adding 1 to both sides, the equation can be written as:

$\tan 4x=1$

In order to solve $x$ we have to replace 1 to tan(π/4) (note: 1 = tan(π/4)), so that we can write 1 in terms of $\tan$ and we'd get:

$\tan 4x=\tan \dfrac{\pi}{4}$

It follows that

$4x=\dfrac{\pi}{4}$

$\therefore x=\dfrac{\frac{\pi}{4}}{4}=\boxed{\dfrac{\pi }{16}}$

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Solving for 2. $\sec^2x-1=0$

We add 1 to both sides to simplify the equation into:

$\sec^2x = 1$

This is just an equation of the form $a^2=c$ and we can solve it by taking square root on both sides. Apply square root on both sides, we have:

$\sqrt{\sec^2x}=\pm\sqrt{1}$

$\sec x=\pm 1$

We know that $\sec$ is the reciprocal of $\cos .$ Therefore,

$\dfrac{1}{\cos x}=\pm 1$

$\cos x=\pm 1$

$\therefore x=\arccos (\pm 1)=\boxed{0, \pi}$

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Solving for 3. $\cos 2x + 3 = 5\cos x$

By the double angle identity of cosine, we have $\cos 2x=\cos^2 x-\sin^2x.$ Substitute this, we get

$\cos ^2x-\sin ^2x+3=5\cos x$

$\cos ^2- \sin^2x + 1 + 2 = 5\cos x$

Now, by Pythagorean Identity, $\sin ^2x + \cos^2x=1.$ Set 1 into $\sin^2x + \cos^2x$ :

$\cos^2x-\sin^2x + (\sin^2x + \cos^2x)+2=5\cos x$

$2 \cos^2x + 2 = 5\cos x$

$2\cos ^2x - 5\cos x + 2=0$

Let $a=\cos x.$ Then

$2a^2-5a+2=0$

$(a-2)(2a-1)=0$

Using the zero product property, we yield

$\begin{array}{ccc} a-2=0, & & 2a-1=0 \\ a=2, & & a=\dfrac{1}{2}\end{array}$

Substitute $a=\cos x$ back, we have one of the solutions $\cos x =2$ , but this is impossible and has no solution because the range of $\cos$ is [-1, 1]. We have the other solution:

$\cos x=\dfrac{1}{2}$

Isolate $x$ by taking the inverse of cosine to both sides:

$\therefore x=\arccos \dfrac{1}{2}=\boxed{\dfrac{\pi}{3}}$

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morillolenzss may recent q pa ko, same topic din. baka kaya mo.

morillolenzss salamat. kaninang umaga pa ko nagsasagot. so far, out of ten, five pa lang nagawa ko. eh due na lang until today. huhu, sana masagot mo pa yunh recent q ko. 1-3 rin.

devoured Due na ba ngayong 13/01?