$\frac{1}{cos θ} + 3 sin θ tan θ + 4 = 0$ $1/costheta+3sinthetatantheta+4=0$ can be expressed as $3 {cos}^{2} θ - 4 cos θ - 4 = 0$ $3cos^2theta - 4costheta -4=0$ . Hence solve the equation $\frac{1}{cos θ} + 3 sin θ tan θ + 4 = 0$ $1/costheta+3sinthetatantheta+4=0$ for $0 \leq θ \leq 360$ $0<=theta<=360$ ?

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$\frac{1}{cos θ} + 3 sin θ tan θ + 4 = 0$ $1/costheta+3sinthetatantheta+4=0$ can be expressed as $3 {cos}^{2} θ - 4 cos θ - 4 = 0$ $3cos^2theta - 4costheta -4=0$ . Hence solve the equation $\frac{1}{cos θ} + 3 sin θ tan θ + 4 = 0$ $1/costheta+3sinthetatantheta+4=0$ for $0 \leq θ \leq 360$ $0<=theta<=360$ ?

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Answer 1 · 2017-11-23T14:11:18

$x = {131.8}^{\circ} or x = {228.2}^{\circ}$ $x=131.8^@" or "x=228.2^@$

Explanation:

$\frac{1}{cos θ} + 3 sin θ \times \frac{sin θ}{cos θ} + 4 = 0$ $1/costheta+3sinthetaxxsintheta/costheta+4=0$

$\Rightarrow \frac{1}{cos θ} + \frac{3 {sin}^{2} θ}{cos θ} + \frac{4 cos θ}{cos θ} = 0$ $rArr1/costheta+(3sin^2theta)/costheta+(4costheta)/costheta=0$

$\Rightarrow \frac{1 + 3 (1 - {cos}^{2} θ) + 4 cos θ}{cos θ} = 0$ $rArr(1+3(1-cos^2theta)+4costheta)/costheta=0$

$\Rightarrow \frac{- 3 {cos}^{2} θ + 4 cos θ + 4}{cos θ} = 0$ $rArr(-3cos^2theta+4costheta+4)/costheta=0"$

$\Rightarrow - 3 {cos}^{2} θ + 4 cos θ + 4 = 0 \leftarrow multiply by - 1$ $rArr-3cos^2theta+4costheta+4=0larrcolor(blue)"multiply by - 1"$

$\Rightarrow 3 {cos}^{2} θ - 4 cos θ - 4 = 0$ $rArr3cos^2theta-4costheta-4=0$

$using 3 {cos}^{2} θ - 4 cos θ - 4 = 0 to solve$ $"using "3cos^2theta-4costheta-4=0" to solve"$

$3 {cos}^{2} θ + 2 cos θ - 6 cos θ - 4 = 0 \leftarrow split cos θ term$ $3cos^2theta+2costheta-6costheta-4=0larr"split "costheta" term"$

$cos θ (3 cos θ + 2) - 2 (3 cos θ + 2) = 0 \leftarrow grouping$ $color(red)(costheta)(3costheta+2)color(red)(-2)(3costheta+2)=0larr"grouping"$

$\Rightarrow (cos θ + 2) (3 cos θ + 2) = 0$ $rArr(costheta+2)(color(red)(3costheta+2))=0$

$equate each factor to zero and solve for θ$ $"equate each factor to zero and solve for "theta$

$cos θ + 2 = 0 \to cos θ = - 2 \leftarrow no solution$ $costheta+2=0tocostheta=-2larrcolor(red)"no solution"$

$3 cos θ + 2 = 0 \to cos θ = - \frac{2}{3}$ $3costheta+2=0tocostheta=-2/3$

$cos θ < 0 \Rightarrow θ in second/third quadrants$ $costheta<0rArrthetacolor(blue)" in second/third quadrants"$

$θ = {cos}^{- 1} (\frac{2}{3}) = {48.2}^{\circ} \leftarrow related acute angle$ $theta=cos^-1(2/3)=48.2^@larrcolor(red)"related acute angle"$

$\Rightarrow θ = {(180 - 48.2)}^{\circ} = {131.8}^{\circ}$ $rArrtheta=(180-48.2)^@=131.8^@$

$\Rightarrow θ = {(180 + 48.2)}^{\circ} = {228.2}^{\circ}$ $rArrtheta=(180+48.2)^@=228.2^@$

Answer 2 · 2017-11-23T14:25:45

$\frac{1}{cos θ} + 3 sin θ tan θ + 4 = 0$ $1/costheta+3sinthetatantheta+4=0$

$\Rightarrow \frac{1}{cos θ} + 3 sin θ \cdot \frac{sin θ}{cos θ} + 4 = 0$ $=>1/costheta+3sintheta*sintheta/costheta+4=0$

$\Rightarrow 1 + 3 {sin}^{2} θ + 4 cos θ = 0$ $=>1+3sin^2theta+4costheta=0$

$\Rightarrow 1 + 3 (1 - {cos}^{2} θ) + 4 cos θ = 0$ $=>1+3(1-cos^2theta)+4costheta=0$

$\Rightarrow 3 {cos}^{2} θ - 4 cos θ - 4 = 0$ $=>3cos^2theta-4costheta-4=0$

$\Rightarrow 3 {cos}^{2} θ - 6 cos θ + 2 cos θ - 4 = 0$ $=>3cos^2theta-6costheta+2costheta-4=0$

$\Rightarrow 3 cos θ (cos θ - 2) + 2 (cos θ - 2) = 0$ $=>3costheta(costheta-2)+2(costheta-2)=0$

$\Rightarrow (3 cos θ + 2) (cos θ - 2) = 0$ $=>(3costheta+2)(costheta-2)=0$

$cos θ = 2$ $costheta=2$ not possible

So $cos θ = - \frac{2}{3} = {cos 131.8}^{\circ} = {cos (360 - 131.8)}^{\circ}$ $costheta=-2/3=cos131.8^@=cos(360-131.8)^@$

Hence

$θ = {131.8}^{\circ} and θ = {228.2}^{\circ}$ $theta=131.8^@ and theta=228.2^@$

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1cosθ+3sinθtanθ+4=01/costheta+3sinthetatantheta+4=0 can be expressed as 3cos2θ−4cosθ−4=03cos^2theta - 4costheta -4=0. Hence solve the equation 1cosθ+3sinθtanθ+4=01/costheta+3sinthetatantheta+4=0 for 0≤θ≤3600<=theta<=360 ?

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