{"id":247280,"date":"2025-07-07T15:41:02","date_gmt":"2025-07-07T15:41:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=247280"},"modified":"2025-07-07T15:41:04","modified_gmt":"2025-07-07T15:41:04","slug":"find-all-solutions-for-arccosx-arcsinx-pi-6","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/find-all-solutions-for-arccosx-arcsinx-pi-6\/","title":{"rendered":"Find all solutions for arccos(x) – arcsin(x) = pi\/6."},"content":{"rendered":"\n

Find all solutions for arccos(x) – arcsin(x) = pi\/6.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are tasked with finding all solutions for the equation:arccos\u2061(x)\u2212arcsin\u2061(x)=\u03c06\\arccos(x) – \\arcsin(x) = \\frac{\\pi}{6}arccos(x)\u2212arcsin(x)=6\u03c0\u200b<\/p>\n\n\n\n

Step 1: Use a trigonometric identity<\/h3>\n\n\n\n

We start by using the identity for inverse trigonometric functions. Specifically, we know that:arccos\u2061(x)+arcsin\u2061(x)=\u03c02\\arccos(x) + \\arcsin(x) = \\frac{\\pi}{2}arccos(x)+arcsin(x)=2\u03c0\u200b<\/p>\n\n\n\n

This identity holds for any value of xxx within the domain [\u22121,1][-1, 1][\u22121,1], since the arccosine and arcsine functions are related in this way.<\/p>\n\n\n\n

Step 2: Express arccos\u2061(x)\\arccos(x)arccos(x) in terms of arcsin\u2061(x)\\arcsin(x)arcsin(x)<\/h3>\n\n\n\n

From the identity above, we can solve for arccos\u2061(x)\\arccos(x)arccos(x):arccos\u2061(x)=\u03c02\u2212arcsin\u2061(x)\\arccos(x) = \\frac{\\pi}{2} – \\arcsin(x)arccos(x)=2\u03c0\u200b\u2212arcsin(x)<\/p>\n\n\n\n

Step 3: Substitute into the given equation<\/h3>\n\n\n\n

Now, substitute arccos\u2061(x)=\u03c02\u2212arcsin\u2061(x)\\arccos(x) = \\frac{\\pi}{2} – \\arcsin(x)arccos(x)=2\u03c0\u200b\u2212arcsin(x) into the original equation:(\u03c02\u2212arcsin\u2061(x))\u2212arcsin\u2061(x)=\u03c06\\left( \\frac{\\pi}{2} – \\arcsin(x) \\right) – \\arcsin(x) = \\frac{\\pi}{6}(2\u03c0\u200b\u2212arcsin(x))\u2212arcsin(x)=6\u03c0\u200b<\/p>\n\n\n\n

Simplify this equation:\u03c02\u22122arcsin\u2061(x)=\u03c06\\frac{\\pi}{2} – 2\\arcsin(x) = \\frac{\\pi}{6}2\u03c0\u200b\u22122arcsin(x)=6\u03c0\u200b<\/p>\n\n\n\n

Step 4: Solve for arcsin\u2061(x)\\arcsin(x)arcsin(x)<\/h3>\n\n\n\n

Now, isolate the term involving arcsin\u2061(x)\\arcsin(x)arcsin(x):\u22122arcsin\u2061(x)=\u03c06\u2212\u03c02-2\\arcsin(x) = \\frac{\\pi}{6} – \\frac{\\pi}{2}\u22122arcsin(x)=6\u03c0\u200b\u22122\u03c0\u200b<\/p>\n\n\n\n

First, simplify the right side:\u03c06\u2212\u03c02=\u03c06\u22123\u03c06=\u22122\u03c06=\u2212\u03c03\\frac{\\pi}{6} – \\frac{\\pi}{2} = \\frac{\\pi}{6} – \\frac{3\\pi}{6} = -\\frac{2\\pi}{6} = -\\frac{\\pi}{3}6\u03c0\u200b\u22122\u03c0\u200b=6\u03c0\u200b\u221263\u03c0\u200b=\u221262\u03c0\u200b=\u22123\u03c0\u200b<\/p>\n\n\n\n

Now the equation becomes:\u22122arcsin\u2061(x)=\u2212\u03c03-2\\arcsin(x) = -\\frac{\\pi}{3}\u22122arcsin(x)=\u22123\u03c0\u200b<\/p>\n\n\n\n

Dividing both sides by -2:arcsin\u2061(x)=\u03c06\\arcsin(x) = \\frac{\\pi}{6}arcsin(x)=6\u03c0\u200b<\/p>\n\n\n\n

Step 5: Solve for xxx<\/h3>\n\n\n\n

Finally, solve for xxx by taking the sine of both sides:x=sin\u2061(\u03c06)x = \\sin\\left(\\frac{\\pi}{6}\\right)x=sin(6\u03c0\u200b)<\/p>\n\n\n\n

Since sin\u2061(\u03c06)=12\\sin\\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}sin(6\u03c0\u200b)=21\u200b, we have:x=12x = \\frac{1}{2}x=21\u200b<\/p>\n\n\n\n

Step 6: Verify the solution<\/h3>\n\n\n\n

Let’s verify the solution by checking that x=12x = \\frac{1}{2}x=21\u200b satisfies the original equation.<\/p>\n\n\n\n

    \n
  1. arccos\u2061(12)=\u03c03\\arccos\\left(\\frac{1}{2}\\right) = \\frac{\\pi}{3}arccos(21\u200b)=3\u03c0\u200b<\/li>\n\n\n\n
  2. arcsin\u2061(12)=\u03c06\\arcsin\\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6}arcsin(21\u200b)=6\u03c0\u200b<\/li>\n<\/ol>\n\n\n\n

    Substituting these into the original equation:arccos\u2061(12)\u2212arcsin\u2061(12)=\u03c03\u2212\u03c06=\u03c06\\arccos\\left(\\frac{1}{2}\\right) – \\arcsin\\left(\\frac{1}{2}\\right) = \\frac{\\pi}{3} – \\frac{\\pi}{6} = \\frac{\\pi}{6}arccos(21\u200b)\u2212arcsin(21\u200b)=3\u03c0\u200b\u22126\u03c0\u200b=6\u03c0\u200b<\/p>\n\n\n\n

    This is true, so the solution is correct.<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    The solution to the equation arccos\u2061(x)\u2212arcsin\u2061(x)=\u03c06\\arccos(x) – \\arcsin(x) = \\frac{\\pi}{6}arccos(x)\u2212arcsin(x)=6\u03c0\u200b is:x=12x = \\frac{1}{2}x=21\u200b<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Find all solutions for arccos(x) – arcsin(x) = pi\/6. The Correct Answer and Explanation is: We are tasked with finding all solutions for the equation:arccos\u2061(x)\u2212arcsin\u2061(x)=\u03c06\\arccos(x) – \\arcsin(x) = \\frac{\\pi}{6}arccos(x)\u2212arcsin(x)=6\u03c0\u200b Step 1: Use a trigonometric identity We start by using the identity for inverse trigonometric functions. Specifically, we know that:arccos\u2061(x)+arcsin\u2061(x)=\u03c02\\arccos(x) + \\arcsin(x) = \\frac{\\pi}{2}arccos(x)+arcsin(x)=2\u03c0\u200b This identity […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-247280","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247280","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=247280"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247280\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=247280"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=247280"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=247280"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}