Solve the following equation for $$ 0^{\circ}

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Solve the equation $$ 3 \cos(x)-2=-1 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$3\cos\left(xight)-2=-1$$
- step1: Move the constant to the right side:
  $$3\cos\left(xight)=-1+2$$
- step2: Add the numbers:
  $$3\cos\left(xight)=1$$
- step3: Divide both sides:
  $$\frac{3\cos\left(xight)}{3}=\frac{1}{3}$$
- step4: Divide the numbers:
  $$\cos\left(xight)=\frac{1}{3}$$
- step5: Use the inverse trigonometric function:
  $$x=\arccos\left(\frac{1}{3}ight)$$
- step6: Calculate:
  $$\begin{align}&x=-\arccos\left(\frac{1}{3}ight)\&x=\arccos\left(\frac{1}{3}ight)\end{align}$$
- step7: Add the period:
  $$\begin{align}&x=-\arccos\left(\frac{1}{3}ight)+2k\pi ,k \in \mathbb{Z}\&x=\arccos\left(\frac{1}{3}ight)+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step8: Find the union:
  $$x=\left\{ \begin{array}{l}-\arccos\left(\frac{1}{3}ight)+2k\pi \\arccos\left(\frac{1}{3}ight)+2k\pi \end{array}ight.,k \in \mathbb{Z}$$
Solve the equation $$ 4 \cot(x-20)-5=10 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$4\cot\left(x-20ight)-5=10$$
- step1: Find the domain:
  $$4\cot\left(x-20ight)-5=10,x
eq 20+k\pi ,k \in \mathbb{Z}$$
- step2: Move the constant to the right side:
  $$4\cot\left(x-20ight)=10+5$$
- step3: Add the numbers:
  $$4\cot\left(x-20ight)=15$$
- step4: Divide both sides:
  $$\frac{4\cot\left(x-20ight)}{4}=\frac{15}{4}$$
- step5: Divide the numbers:
  $$\cot\left(x-20ight)=\frac{15}{4}$$
- step6: Use the inverse trigonometric function:
  $$x-20=\operatorname{arccot}\left(\frac{15}{4}ight)$$
- step7: Add the period:
  $$x-20=\operatorname{arccot}\left(\frac{15}{4}ight)+k\pi ,k \in \mathbb{Z}$$
- step8: Solve the equation:
  $$x=\operatorname{arccot}\left(\frac{15}{4}ight)+20+k\pi ,k \in \mathbb{Z}$$
- step9: Check if the solution is in the defined range:
  $$x=\operatorname{arccot}\left(\frac{15}{4}ight)+20+k\pi ,k \in \mathbb{Z},x
eq 20+k\pi ,k \in \mathbb{Z}$$
- step10: Find the intersection:
  $$x=\operatorname{arccot}\left(\frac{15}{4}ight)+20+k\pi ,k \in \mathbb{Z}$$
The solutions to the given equations are:

1. For the equation $$3 \cos(x) - 2 = -1$$, the solutions are $$x = -\arccos(\frac{1}{3}) + 2k\pi$$ and $$x = \arccos(\frac{1}{3}) + 2k\pi$$, where $$k$$ is an integer.

2. For the equation $$4 \cot(x-20) - 5 = 10$$, the solutions are $$x = \arccot(\frac{15}{4}) + 20 + k\pi$$, where $$k$$ is an integer.

These solutions are valid for the given range $$0^{\circ} < x < 90^{\circ}$$.
The solutions for the equations are: 1. $$x = \arccos\left(\frac{1}{3}ight)$$ 2. $$x = \arccot\left(\frac{15}{4}ight) + 20^{\circ}$$ These solutions are valid within the range $$0^{\circ} < x < 90^{\circ}$$.

Solve the following equation for \( 0^{\circ}

Ask by Dawson Ross. in South Africa

Mar 12,2025

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Solve the following equation for \( 0^{\circ} Ask by Dawson Ross. in South Africa Mar 12,2025