Lesson Miscellaneous problems on solving trigonometric equations
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<H2>Miscellaneous problems on solving trigonometric equations</H2> <H3>Problem 1</H3>Solve an equation sin(3x) + cos(3x) = -1. <B>Solution</B> <pre> The original equation is sin(3x) + cos(3x) = -1. (1) Square its both sides. You will get {{{sin^2(3x) + 2*sin(3x)*cos(3x) + cos^2(3x)}}} = {{{1}}}. (2) From the other side, there is an identity {{{sin^2(3x) + cos^2(3x)}}} == {{{1}}}. (3) Comparing (2) and (3), you get 2*sin(3x)*cos(3x) = 0, or sin(3x)*cos(3x) = 0. (4) Equation (4) splits in two independent equations 1) sin(3x) = 0 ---> 3x = {{{k*pi}}}, k = 0. +/-1. +/-2, . . . ---> x = {{{(k*pi)/3}}}, k = 0, +/-1. +/-2, . . . (5) 2) cos(3x) = 0 ---> 3x = {{{pi/2 + k*pi}}}, k = 0. +/-1. +/-2, . . . ---> x = {{{pi/6 + (k*pi)/3}}}, k = 0. +/-1. +/-2, . . . (6) Now we should check which of the found values (5), (6) satisfy the original equation. Of the set (5), all x satisfy sin(3x) = 0. Hence, only those of (5) satisfy the original equation where cos(3x) = -1. They are 3x = {{{pi + 2n*pi}}}, n = 0, +/-1. +/-2, . . . , or x = {{{((2n+1)/3)*pi}}}, n = 0, +/-1. +/-2, . . . , (5'). Of the set (6), all x satisfy cos(3x) = 0. Hence, only those of (6) satisfy the original equation where sin(3x) = -1. They are 3x = {{{3pi/2 + 2n*pi}}}, n = 0, +/-1. +/-2, . . . , or x = {{{(1/2+(2n)/3)*pi}}}, n = 0, +/-1. +/-2, . . . , (6'). <U>Answer</U>. The union of the sets (5') and (6') is the solution of the original equation. </pre> <H3>Problem 2</H3>Solve an equation cos(6x) = cos(x) for 0 <= x < 360 degrees <B>Solution</B> <pre> Since cos(6x) = cos(x), it implies one of two possibilities: (1) EITHER 6x = x + 360n degrees (2) OR 6x = -x + 360n degrees. From (1), we have 6x-x = 360n; 5x = 360n; y = 0, 360/5 = 72, (360/5)*2, . . . , (360/5)*5. Taking into account the restriction on the range, the only possible solutions are 0°, 72°, 144°, 216°, 288°. From (2), we have 6x+x = 360n; 7x = 360n; x = 0, 360/7, (360/7)*2, (360/7)*3, (360/7)*4, (360/7)*5, (360/7)*6. Taking into account the restriction on the range, the only possible solutions are 0° and 360/7 degrees. It gives the <U>ANSWER</U> : the solutions for x are 0°, 72°, 144°, 216°, 288°, 360/7, (360/7)*2, (360/7)*3, (360/7)*4, (360/7)*5, (360/7)*6 degrees. </pre> <H3>Problem 3</H3>Find the solutions of equation {{{tan(x+pi)}}} − {{{tan(pi-x)}}} = 0 for 0 <= x < {{{2pi}}}. <B>Solution</B> <pre> Your starting equation is {{{tan(x+pi)}}} − {{{tan(pi-x)}}} = 0 for 0 <= x < {{{2pi}}}. It is the same as (is equivalent to) {{{tan(x+pi)}}} = {{{tan(pi-x)}}} for 0 <= x < {{{2pi}}}. Since tan is a monotonic periodic function with the period of π, it means that {{{(x+pi)}}} - {{{(pi-x)}}} = {{{k*pi}}}, where k = 0, +/-1, +/-2, . . . or 2x = {{{k*pi}}}, where k = 0, +/-1, +/-2, . . . Having x in the interval 0 <= x < {{{2*pi}}}, it means that x = 0 or x = {{{pi}}}. <U>ANSWER</U> </pre> <H3>Problem 4</H3>Find the values of x, where 0° < x < 180° such that 2cos^2(x) + sin(20°) = 1. <B>Solution</B> <pre> Your starting equation is 2cos^2(x) + sin(20°) = 1. (1) Notice that sin(20°) = cos(90°-20°) = cos(70°). Therefore, you can re-write the original equation in the form 2cos^2(x) = 1 - cos(70°), and further as cos^2(x) = {{{(1-cos(70^o))/2}}}. (2) Next, compare it with the formula for sine of the half angle {{{sin^2(a/2)}}} = {{{(1-cos^2(a))/2)}}}. Taking a = 70°, you see from (2) that {{{(1-cos(70^o))/2}}} = {{{sin^2(35^o)}}}. (3) From (2) and (3), we have {{{cos^2(x)}}} = {{{sin^2(35^o)}}}. (4) Now use sin(35°) = cos(90°-35°) = cos(55°). Then equation (4) takes the form {{{cos^2(x)}}} = {{{cos^2(55^o)}}}. (5) It gives for x the following possibilities x = 55°, 180°-55° = 125°, 55°+180° = 235° and 360°-55° = 305°. Since we are looking for the solutions in the interval 0° < x < 180° only, it leaves only two possible values. <U>ANSWER</U>. In the given interval, the solutions are x= 55° and x= 125°. <U>CHECK</U>. cos(55°) = 0.573576; sin(20°) = 0.342020; 2*0.573576^2 + 0.342020 = 0.999999. ! 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