Question

solve the equation algebraically, and then use a calculator to find the values on the interval $0,2\pi)$. round answers to 4 decimal places.

$\tan^{2}(x)+3\tan(x)-3 = 0$

$x=$

Explanation:

Step1: Let \( t = \tan(x) \)

The equation \( \tan^{2}(x)+3\tan(x) - 3=0 \) becomes \( t^{2}+3t - 3=0 \)

Step2: Use the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)

For \( t^{2}+3t - 3=0 \), where \( a = 1 \), \( b=3 \), \( c=-3 \)

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Step3: Calculate the values of \( t \)

\( t_{1}=\frac{-3+\sqrt{21}}{2}\approx\frac{-3 + 4.5837}{2}=\frac{1.5837}{2}=0.7919 \)
\( t_{2}=\frac{-3-\sqrt{21}}{2}\approx\frac{-3-4.5837}{2}=\frac{-7.5837}{2}=-3.7919 \)

Step4: Solve for \( x \) when \( \tan(x)=t \)

When \( \tan(x)=0.7919 \), \( x=\arctan(0.7919)\approx0.6662 \) and \( x=\pi+ 0.6662\approx3.8078 \)
When \( \tan(x)=-3.7919 \), \( x=\arctan(-3.7919)+\pi\approx -1.3289+\pi\approx1.8127 \) and \( x=\arctan(-3.7919)+2\pi\approx -1.3289 + 2\pi\approx4.9545 \)

Answer:

\( 0.6662,1.8127,3.8078,4.9545 \)